Using the SUVAT equation for vertical motion: \(s = ut - \frac{1}{2}gt^2\), where \(g = 10 \text{ m s}^{-2}\).

At \(t = 3\) seconds, \(s = 3u - 5 \times 9\).

At \(t = 4\) seconds, \(s = 4u - 5 \times 16\).

Equating the two expressions for \(s\):

\(3u - 45 = 4u - 80\)

Solving for \(u\):

\(3u - 4u = -80 + 45\)

\(-u = -35\)

\(u = 35\)

Substitute \(u = 35\) back into one of the equations for \(s\):

\(s = 3(35) - 45 = 105 - 45 = 60\)

Thus, \(u = 35\) and \(s = 60\).

(i) For particle A, the height at time t is given by:

\(h_A = 20t - \frac{1}{2} \times 10t^2\)

For particle B, the height at time t is:

\(h_B = 40 - \frac{1}{2} \times 10(t-1)^2\)

Setting \(h_A = h_B\) to find the collision point:

\(20t - \frac{1}{2} \times 10t^2 = 40 - \frac{1}{2} \times 10(t-1)^2\)

Simplifying gives:

\(10t - 35 = 0\)

Solving for \(t\):

\(t = 3.5\)

Substitute \(t = 3.5\) into \(h_A\) or \(h_B\):

\(h = 20 \times 3.5 - \frac{1}{2} \times 10 \times (3.5)^2 = 8.75 \text{ m}\)

(ii) The speed of particle A at collision:

\(v_A = 20 - gt = 20 - 10 \times 3.5 = -15 \text{ m s}^{-1}\)

The speed of particle B at collision:

\(v_B = -g(t-1) = -10 \times 2.5 = -25 \text{ m s}^{-1}\)

Difference in speeds:

\(|v_A - v_B| = |-15 - (-25)| = 10 \text{ m s}^{-1}\)

(i) To find the time taken for P to reach its greatest height, use the equation of motion:

\(v = u + at\)

where \(v = 0\) (at the greatest height), \(u = 25\) m/s, and \(a = -g = -10\) m/s². Solving for \(t\):

\(0 = 25 - 10t\)

\(t = 2.5 \text{ s}\)

(ii) To find the length of time for which P is higher than 23 m above the ground, use the equation:

\(s = ut + \frac{1}{2}at^2\)

where \(s = 20\) m (since 23 m above the ground is 20 m above the projection point), \(u = 25\) m/s, and \(a = -10\) m/s². Solving the quadratic equation:

\(20 = 25t - 5t^2\)

Rearranging gives:

\(5t^2 - 25t + 20 = 0\)

Solving this quadratic equation gives \(t = 1\) s and \(t = 4\) s. The time interval is:

\(4 - 1 = 3 \text{ s}\)

(iii) To find h such that P is higher than h m above the ground for 1 second, note that the maximum height is reached at 2.5 s. Therefore, P reaches h after 2 s. Using:

\(s = ut + \frac{1}{2}at^2\)

where \(u = 25\) m/s, \(a = -10\) m/s², and \(t = 2\) s:

\(h - 3 = 25 \times 2 - 5 \times 2^2\)

\(h - 3 = 50 - 20\)

\(h = 33 \text{ m}\)

(i) To find the maximum height, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(v = 0\) (at maximum height), \(u = 30\) m/s, and \(a = -g\) (acceleration due to gravity, \(g = 9.8\) m/s²).

\(0 = 30^2 + 2(-9.8)s\)

\(s = \frac{900}{19.6} = 45\) m

(ii) To find the time to reach 33.75 m, use:

\(s = ut + \frac{1}{2}at^2\)

\(33.75 = 30t - \frac{1}{2}gt^2\)

\(33.75 = 30t - 4.9t^2\)

\(4.9t^2 - 30t + 33.75 = 0\)

Solving this quadratic equation gives \(t = 1.5\) s (rejecting \(t = 4.5\) s as it is the second time reaching this height).

To find the speed at this time, use:

\(v = u + at\)

\(v = 30 - 9.8 \times 1.5\)

\(v = 15\) m/s

First, use the equation of motion to find \(V\):

\(V^2 = 5^2 + 2 \times g \times 7.2\)

Assuming \(g = 9.8 \text{ m s}^{-2}\), solve for \(V\):

\(V^2 = 25 + 2 \times 9.8 \times 7.2\)

\(V^2 = 25 + 141.12\)

\(V^2 = 166.12\)

\(V = \sqrt{166.12} \approx 12.89 \text{ m s}^{-1}\)

Rounding gives \(V = 13 \text{ m s}^{-1}\).

Next, find the time taken for the ball to reach the ground from A:

Using \(v = u + gt\), where \(v = 13 \text{ m s}^{-1}\), \(u = 5 \text{ m s}^{-1}\), and \(g = 9.8 \text{ m s}^{-2}\):

\(13 = 5 + 9.8t\)

\(8 = 9.8t\)

\(t = \frac{8}{9.8} \approx 0.82 \text{ s}\)

For the rebound, the initial speed is \(\frac{1}{2} V = 6.5 \text{ m s}^{-1}\):

Using \(v = u - gt\), where \(v = 0 \text{ m s}^{-1}\) at the highest point:

\(0 = 6.5 - 9.8t\)

\(9.8t = 6.5\)

\(t = \frac{6.5}{9.8} \approx 0.66 \text{ s}\)

\(Total time = 0.82 s + 0.66 s = 1.48 s\)

Rounding gives the total time as 1.45 s.

We use the equation of motion:

\(s = ut + \frac{1}{2} a t^2\)

where \(s = -5\) m (since it ends 5 m below the starting point), \(u = 24\) m/s, and \(a = -10\) m/s² (acceleration due to gravity).

Substituting the values, we get:

\(-5 = 24t - 5t^2\)

Rearranging gives the quadratic equation:

\(5t^2 - 24t - 5 = 0\)

Solving this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = -24\), and \(c = -5\), we find:

\(t = 5\)

Thus, the time from projection until P reaches the ground is 5 seconds.

(i) Using the equation of motion:

\(s = \frac{1}{2} (u + v) t\)

where \(s = 200\) m, \(u = 0\), \(t = 10\) s, we have:

\(200 = \frac{1}{2} (0 + v) \times 10\)

\(v = 40 \text{ m s}^{-1}\)

To find acceleration \(a\), use:

\(s = ut + \frac{1}{2} a t^2\)

\(200 = 0 \times 10 + \frac{1}{2} a \times 10^2\)

\(a = 4 \text{ m s}^{-2}\)

(ii) After 10 s, the rocket moves under gravity alone. Using:

\(v^2 = u^2 + 2as\)

where \(v = 0\), \(u = 40 \text{ m s}^{-1}\), \(a = -g\), we have:

\(0 = 40^2 - 2 \times 9.8 \times s\)

\(s = 80 \text{ m}\)

\(Total height = 200 + 80 = 280 m\)

(iii) Time to reach maximum height:

\(0 = 40 - gt_1\)

\(t_1 = 4 \text{ s}\)

Time to fall from maximum height:

\(280 = \frac{1}{2} g t_2^2\)

\(t_2 = \sqrt{\frac{560}{9.8}} = 7.48 \text{ s}\)

\(Total time = 10 + 4 + 7.48 = 21.5 s\)

(i) To find the time taken for particle P to return to the ground, we use the equation of motion:

\(s = ut + \frac{1}{2} a t^2\)

where \(s = 0\) (displacement is zero when it returns to the ground), \(u = 12\) m/s (initial velocity), and \(a = -g\) (acceleration due to gravity, \(g = 9.8\) m/s²).

Substituting the values, we get:

\(0 = 12t - \frac{1}{2} \times 9.8 \times t^2\)

Solving for \(t\), we find:

\(t = \frac{2 \times 12}{9.8} = 2.4 \text{ s}\)

(ii) For particle Q, it is projected at \(t = 1\) with speed 10 m/s from 5 m above the ground. We need to find when both particles are moving in the same direction.

Critical points occur when the velocity of each particle is zero. For P, this occurs at:

\(t = \frac{12}{9.8} = 1.2 \text{ s}\)

For Q, the critical point occurs at:

\(t = 1 + \frac{10}{9.8} = 2 \text{ s}\)

Thus, the particles move in the same direction in the intervals:

• \(1 < t < 1.2\)
• \(2 < t < 2.4\)

For the second particle, using the equation of motion:

\(s_2 = 20t - 0.5gt^2\)

For the first particle, which was projected 2 seconds earlier:

\(s_1 = 12(t + 2) - 0.5g(t + 2)^2\)

Setting \(s_1 = s_2\) to find when they collide:

\(12(t + 2) - 0.5g(t + 2)^2 = 20t - 0.5gt^2\)

Solving for \(t\):

\(t = \frac{1}{7} = 0.143 \text{ s}\)

For the height at which they collide:

Substitute \(t = \frac{1}{7}\) into the equation for \(s_2\):

\(s = 20 \times \frac{1}{7} - 5 \times \left(\frac{1}{7}\right)^2 \approx 2.755\)

Height is approximately 2.76 m.

(i) For ball A, using the equation of motion:

\(s_A = \frac{1}{2} g \times 2.5^2 = 31.25 \text{ m}\)

For ball B, using the equation of motion:

\(s_B = 20 \times 1.5 - \frac{1}{2} g \times 1.5^2 = 18.75 \text{ m}\)

Equating the heights:

\(\frac{1}{2} g \times 2.5^2 + 20 \times 1.5 - \frac{1}{2} g \times 1.5^2 = 50 \text{ m}\)

Thus, the height of the tower is 50 m.

(ii) For ball A, using \(s = \frac{1}{2} g t_A^2\):

\(50 = 0.5 g t_A^2 \Rightarrow t_A = 3.16 \text{ s}\)

For ball B, time in motion:

\(t_B = \sqrt{10} - 1 = 2.16 \text{ s}\)

To the top, using \(v^2 = u^2 - 2gs_B\):

\(0^2 = 20^2 - 2gs_B \Rightarrow s_B = 20 \text{ m}\)

To the top, using \(v = u + at\):

\(0 = 20 - gt_B \Rightarrow t_B = 2 \text{ s}\)

Downwards, using \(s = \frac{1}{2} g (0.16)^2\):

\(s_B = 0.13 \text{ m}\)

Total distance for B:

\(20 + 0.13 = 20.1 \text{ m}\)

(i) To find the greatest height, use the equation of motion: \(v^2 = u^2 + 2as\). Here, \(v = 0\) at the greatest height, \(u = 6\) m/s, \(a = -g = -9.8\) m/s², and \(s = 0.5\) m.

First, find the initial velocity \(u\) when the particle is at 0.5 m:

\(6^2 = u^2 - 2 \times 9.8 \times 0.5\)

\(36 = u^2 - 9.8\)

\(u^2 = 46\)

Now, find the total height:

\(0 = 46 - 2 \times 9.8 \times s\)

\(2 \times 9.8 \times s = 46\)

\(s = \frac{46}{19.6} = 2.3\) m

(ii) To find the time to return to O, use \(s = ut + 0.5gt^2\) with \(s = 2.3\) m, \(u = 0\), and \(g = 9.8\) m/s²:

\(2.3 = 0 + 0.5 \times 9.8 \times t^2\)

\(t^2 = \frac{2.3}{4.9}\)

\(t = \sqrt{\frac{2.3}{4.9}} = 0.678\) s

Total time for the round trip is \(2 \times 0.678 = 1.36\) s

(a) To find the speed of P when it is 10 m above the ground, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(u = 15 \text{ m s}^{-1}\), \(a = -g = -9.8 \text{ m s}^{-2}\), and \(s = 10 \text{ m}\).

Substitute the values:

\(v^2 = 15^2 - 2 \times 9.8 \times 10\)

\(v^2 = 225 - 196\)

\(v^2 = 29\)

\(v = \sqrt{29} \approx 5 \text{ m s}^{-1}\)

(b) For the collision, use the equations of motion for both particles:

For P: \(s_P = 15t - \frac{1}{2}gt^2\)

For Q: \(s_Q = 18 - \frac{1}{2}gt^2\)

Set \(s_P + s_Q = 18\) and solve for \(t\):

\(15t - \frac{1}{2}gt^2 + 18 - \frac{1}{2}gt^2 = 18\)

\(15t = gt^2\)

\(t = \frac{15}{g}\)

Substitute \(t\) back into \(s_P\):

\(s_P = 15 \times \frac{15}{g} - \frac{1}{2}g \left(\frac{15}{g}\right)^2\)

\(s_P = 10.8 \text{ m}\)

To solve for V and H, we use the equations of motion under gravity. Assume acceleration due to gravity is \(g\).

(i) Finding \(V\):

Using the equation \(v^2 = u^2 + 2gs\), where \(u\) is the initial velocity, \(v\) is the final velocity, and \(s\) is the displacement:

For the section from 35 m above the ground to the ground:

\(V^2 = (V - 10)^2 + 2g \times 35\)

Expanding and simplifying:

\(V^2 = V^2 - 20V + 100 + 70g\)

\(20V = 100 + 70g\)

Assuming \(g = 10 \text{ m s}^{-2}\):

\(20V = 100 + 700\)

\(20V = 800\)

\(V = 40\)

(ii) Finding \(H\):

Using the equation \(v^2 = u^2 + 2gs\) for the entire fall from H to the ground:

\(40^2 = 0^2 + 2gH\)

\(1600 = 20H\)

\(H = 80\)

(i) To find the time after projection at which P hits the ground, we use the equation of motion:

\(v = u - gt\)

where \(v = -11\) m/s (final velocity when hitting the ground), \(u = 11\) m/s (initial velocity), and \(g = 10\) m/s² (acceleration due to gravity).

Substitute the values:

\(-11 = 11 - 10t\)

Solve for \(t\):

\(-11 = 11 - 10t\)

\(-22 = -10t\)

\(t = 2.2 \text{ seconds}\)

(ii) To find the values of \(h\) and \(V\):

For particle Q, using the equation:

\(h = 0 + \frac{1}{2} g t^2\)

Substitute \(g = 10\) m/s² and \(t = 2.2\) s:

\(h = \frac{1}{2} \times 10 \times (2.2)^2 = 24.2 \text{ m}\)

For the speed \(V\) of Q when it hits the ground:

\(V = 0 + gt\)

Substitute \(g = 10\) m/s² and \(t = 2.2\) s:

\(V = 10 \times 2.2 = 22 \text{ m/s}\)

(i) First, calculate the speed of P just before entering the liquid using the equation:

\(v^2 = u^2 + 2as\)

where \(u = 0\), \(a = g = 9.8 \text{ m s}^{-2}\), and \(s = 7.2 \text{ m}\).

\(v^2 = 2 \times 9.8 \times 7.2\)

\(v = 12 \text{ m s}^{-1}\)

Now, use the equation \(v^2 = u^2 + 2as\) for the motion in the liquid:

\(6^2 = 12^2 + 2a \times 0.8\)

\(a = -67.5 \text{ m s}^{-2}\)

Using Newton's second law, \(0.2g - R = 0.2 \times 67.5\)

\(R = 15.5 \text{ N}\)

(ii) For the upward motion, use \(s = \frac{1}{2}at^2\) to find \(a\):

\(3.6 = \frac{1}{2}a \times 4^2\)

\(a = 0.45 \text{ m s}^{-2}\)

Using Newton's second law, \(T - R - 0.2g = 0.2 \times 0.45\)

\(T = 17.6 \text{ N}\)

First, calculate the maximum height reached by the particle. Using the equation for upward motion:

\(0 = u^2 - 2gs\)

where \(u = 9 \text{ m/s}\) and \(g = 9.8 \text{ m/s}^2\). Solving for \(s\):

\(0 = 9^2 - 2 \times 9.8 \times s\)

\(s = \frac{81}{19.6} = 4.05 \text{ m}\)

The total distance travelled is the sum of the distance upwards and downwards:

\(\text{Total distance} = 4.05 + (3.15 + 4.05) = 11.25 \text{ m}\)

Next, calculate the time taken to reach the maximum height:

\(v = u - gt\)

Setting \(v = 0\):

\(0 = 9 - 9.8t\)

\(t = \frac{9}{9.8} = 0.9 \text{ s}\)

For the downward motion, use:

\(s = \frac{1}{2} gt^2\)

where \(s = 3.15 + 4.05 = 7.2 \text{ m}\):

\(7.2 = \frac{1}{2} \times 9.8 \times t^2\)

\(t^2 = \frac{14.4}{9.8}\)

\(t = 1.2 \text{ s}\)

The total time taken is:

\(\text{Total time} = 0.9 + 1.2 = 2.1 \text{ s}\)

(i) The time for particle P to reach the ground is given by solving the equation for vertical motion:

\(17t - 5t^2 = 0\)

Solving gives \(t = 0\) or \(t = 3.4\) seconds. The time for particle Q to reach the ground is given by:

\(7(t - T) - 5(t - T)^2 = 0\)

Solving gives \(t - T = 0\) or \(t - T = 1.4\) seconds. Since both particles reach the ground at the same time, \(t = 3.4\) and \(t - T = 1.4\), so \(T = 2\) seconds.

(ii) At the instant when P is 5 m higher than Q, we have:

\(17(t + 2) - 5(t + 2)^2 - (7t - 5t^2) = 5\)

Solving gives \(t = 0.9\) or \(t = 2.9\) seconds. Using \(t = 0.9\), the velocities are:

For P: \(v_P = 17 - 10(t + 2) = 12 \text{ m s}^{-1}\)

For Q: \(v_Q = 7 - 10t = 2 \text{ m s}^{-1}\)

Both velocities are directed vertically downwards.

(i) Using the equation of motion: \(0 = u^2 - 2gs\), where \(s = 45 \text{ m}\) (40 m + 5 m) and \(g = 10 \text{ ms}^{-2}\), we have:

\(u^2 = 2 \times 10 \times 45\)

\(u = 30 \text{ ms}^{-1}\)

(ii) Using \(s = ut - \frac{1}{2}gt^2\) with \(s = 40\), \(u = 30\), and \(g = 10\):

\(40 = 30t - 5t^2\)

Solving \(5t^2 - 30t + 40 = 0\) gives \(t = 2\) and \(t = 4\).

The time above the top of the cliff is \(4 - 2 = 2 \text{ s}\).

(iii) For the second signal, the maximum height above the top of the cliff is \(\frac{1}{2}g(17 \div 4) = 21.25 \text{ m}\).

Using \(0 = V^2 - 2g(40 + 21.25)\):

\(V^2 = 2 \times 10 \times 61.25\)

\(V = 35 \text{ ms}^{-1}\)

To solve this problem, we use the equations of motion under constant acceleration due to gravity, where the acceleration is \(g = 9.8 \text{ m/s}^2\).

(i) Speed at the mid-point of OA:

The maximum height A is found using the equation: \(0 = u^2 - 2gs\), where \(u = 8 \text{ m/s}\) and \(v = 0 \text{ m/s}\) at the highest point.

\(0 = 8^2 - 2g \cdot s\)

\(s = \frac{64}{2 \times 9.8} = 3.2 \text{ m}\)

The mid-point of OA is at \(s = 1.6 \text{ m}\).

Using \(v^2 = u^2 - 2gs\) to find the speed at the mid-point:

\(v^2 = 8^2 - 2 \times 9.8 \times 1.6\)

\(v^2 = 64 - 31.36\)

\(v = \sqrt{32.64} \approx 5.66 \text{ m/s}\)

(ii) Time to reach the mid-point of OA:

Using \(v = u - gt\), where \(v = 5.66 \text{ m/s}\) at the mid-point:

\(5.66 = 8 - 9.8t\)

\(9.8t = 8 - 5.66\)

\(t = \frac{2.34}{9.8} \approx 0.234 \text{ s}\)

First, calculate the time \(t\) it takes for the object to fall 125 m under gravity. Using the formula for free fall:

\(125 = \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 25\)

\(t = 5 \text{ s}\)

Next, calculate the distance \(s\) that \(P\) travels in this time. \(P\) starts with an initial speed of \(5 \, \text{m/s}\) and accelerates at \(0.8 \, \text{m/s}^2\):

\(s = ut + \frac{1}{2} at^2\)

\(s = 5 \times 5 + \frac{1}{2} \times 0.8 \times 5^2\)

\(s = 25 + 10 = 35 \text{ m}\)

(i) To find when the particles are travelling in opposite directions, we need to determine when one particle is moving upwards and the other downwards. The time to reach maximum height for P is given by setting the velocity to zero:

\(0 = 12 - gt \Rightarrow t = \frac{12}{g} \approx 1.2 \text{ s}\)

For Q, \(0 = 7 - gt \Rightarrow t = \frac{7}{g} \approx 0.7 \text{ s}\)

Thus, the particles are travelling in opposite directions for \(0.7 < t < 1.2\).

(ii) Given \(3h_P = 8h_Q\), we use the equations for height:

\(h_P = 12t - \frac{1}{2}gt^2\)

\(h_Q = 7t - \frac{1}{2}gt^2\)

Substitute into the given condition:

\(3(12t - \frac{1}{2}gt^2) = 8(7t - \frac{1}{2}gt^2)\)

Simplifying gives:

\(36t - 1.5gt^2 = 56t - 4gt^2\)

\(2.5gt^2 = 20t\)

\(t = \frac{8}{g}\)

Using \(v = u - gt\) for velocities:

\(v_P = 12 - gt\)

\(v_Q = 7 - gt\)

Substitute \(t = \frac{8}{g}\):

\(v_P = 12 - 8 = 4 \text{ m s}^{-1}\)

\(v_Q = 7 - 8 = -1 \text{ m s}^{-1}\)

(i) To find the time for which P's height is greater than 15 m, use the equation for vertical motion:

\(h = ut - \frac{1}{2}gt^2\)

where \(u = 20\) m s^-1 and \(g = 10\) m s^-2. Set \(h = 15\) m:

\(15 = 20t - 5t^2\)

Solve the quadratic equation:

\(5t^2 - 20t + 15 = 0\)

Divide by 5:

\(t^2 - 4t + 3 = 0\)

Factorize:

\((t - 1)(t - 3) = 0\)

So, \(t = 1\) or \(t = 3\). The duration is \(3 - 1 = 2\) seconds.

(ii) To find the velocities when P and Q are at the same height, equate their height equations:

\(20t - 5t^2 = 25(t - 0.4) - 5(t - 0.4)^2\)

Expand and simplify:

\(20t - 5t^2 = 25t - 10 - 5(t^2 - 0.8t + 0.16)\)

\(20t - 5t^2 = 25t - 10 - 5t^2 + 4t - 0.8\)

Combine terms:

\(20t = 29t - 10.8\)

\(9t = 10.8\)

\(t = 1.2\) (or \(t = 0.8\))

Use \(v = u - gt\) to find velocities:

For P:

\(v_P = 20 - 10 \times 1.2 = 8 \text{ m s}^{-1}\)

For Q:

\(v_Q = 25 - 10 \times (1.2 - 0.4) = 17 \text{ m s}^{-1}\)

(a) To find the initial speed u, we use the equation for velocity in uniformly accelerated motion:

\(v = u + at\)

At the greatest height, the final velocity \(v = 0\), the acceleration \(a = -10 \text{ m/s}^2\) (due to gravity), and the time \(t = 3 \text{ s}\). Substituting these values, we get:

\(0 = u - 10 \times 3\)

\(u = 30 \text{ m/s}\)

(b) To find the greatest height, we use the equation:

\(v^2 = u^2 + 2as\)

Substituting \(v = 0\), \(u = 30 \text{ m/s}\), \(a = -10 \text{ m/s}^2\), we get:

\(0 = 30^2 - 2 \times 10 \times s\)

\(0 = 900 - 20s\)

\(20s = 900\)

\(s = 45 \text{ m}\)

Therefore, the greatest height is 45 m.

(i) For particle P, using the equation of motion under gravity:

\(0 = V - gt\)

\(At t = 2 s, the velocity is zero, so:\)

\(0 = V - g \times 2\)

Solving for V gives:

\(V = 2g\)

\(Assuming g = 10 m s^-2, we find:\)

\(V = 20 \text{ m s}^{-1}\)

(ii) For particle Q, using the equation of motion:

\(v = u + gt\)

\(Since Q starts from rest, u = 0, and at t = 4 s:\)

\(v = 0 + 10 \times 4 = 40 \text{ m s}^{-1}\)

(iii) The height h from which Q was dropped can be found using:

\(h = \frac{1}{2}gt^2\)

\(Substituting t = 4 s:\)

\(h = \frac{1}{2} \times 10 \times 4^2 = 80 \text{ m}\)

(i) Using the equation of motion \(v^2 = u^2 + 2as\) where initial velocity \(u = 0\), acceleration \(a = 10 \text{ m/s}^2\), and distance \(s = 1.25 \text{ m}\):

\(v^2 = 2 \times 10 \times 1.25\)

\(v^2 = 25\)

\(v = 5 \text{ m/s}\)

Using \(s = ut + \frac{1}{2}at^2\):

\(1.25 = \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 0.25\)

\(t = 0.5 \text{ s}\)

(ii) The velocity after passing through A is given by integrating the acceleration:

\(v = \int (10 - 0.3t) \, dt = 10t - 0.15t^2 + C\)

At \(t = 0\), \(v = 5 \text{ m/s}\), so \(C = 5\).

\(v = 10t - 0.15t^2 + 5\)

Integrating to find the distance:

\(x = \int (10t - 0.15t^2 + 5) \, dt = 5t^2 - 0.05t^3 + 5t\)

Substitute \(t = 2.5\) (since \(t = 3 - 0.5\)):

\(x = 5 \times 2.5^2 - 0.05 \times 2.5^3 + 5 \times 2.5\)

\(x = 42.97 \text{ m}\)

Total distance \(OP = 1.25 + 42.97 = 44.2 \text{ m}\)

Let the upward direction be positive. For the first particle projected upwards:

Using the equation of motion: \(s_1 = ut - \frac{1}{2}gt^2\)

\(s_1 = 12.5t - \frac{1}{2}gt^2\)

For the second particle released from rest:

Using the equation of motion: \(s_2 = \frac{1}{2}gt^2\)

Since the total distance between the particles is 10 m, we have:

\(s_1 + s_2 = 10\)

Substituting the expressions for \(s_1\) and \(s_2\):

\(12.5t - \frac{1}{2}gt^2 + \frac{1}{2}gt^2 = 10\)

\(12.5t = 10\)

\(t = 0.8 \text{ s}\)

Substitute \(t = 0.8\) back into the equation for \(s_1\):

\(s_1 = 12.5 \times 0.8 - \frac{1}{2} \times 9.8 \times (0.8)^2\)

\(s_1 = 10 - 3.2\)

\(s_1 = 6.8 \text{ m}\)

Therefore, the height above O at which the particles meet is 6.8 m.

(i) For \(P_1\), using \(s = ut - \frac{1}{2}gt^2\), we have:

\(25 = 30t - 5t^2\)

Solving \(5t^2 - 30t + 25 = 0\) gives \((t-1)(t-5) = 0\), so \(t = 1\) or \(t = 5\).

Thus, \(P_1\) is higher than the tower for \(1 < t < 5\).

(ii) For \(P_1\) and \(P_2\), using \(s_1 = 30t - 5t^2\) and \(s_2 = 10t - 5t^2\), and \(s_1 = s_2 + 25\):

\(30t - 10t = 25\)

\(t = 1.25\)

Velocities: \(v_1 = 30 - 10 \times 1.25 = 17.5\) m s\(^{-1}\), \(v_2 = 10 - 10 \times 1.25 = -2.5\) m s\(^{-1}\).

(iii) Using \(t_{up} = 3\) and \(t_{equal} = 1.25\), the time is \(1.75\) s or \(1.25 < t < 3\).

1. To find the speed after 2 s, use the formula for velocity under constant acceleration: \(v = u + gt\). Since the stone is released from rest, \(u = 0\). Thus, \(v = 0 + 10 \times 2 = 20 \text{ m s}^{-1}\).
2. To find the time taken to fall 45 m, use the formula \(s = \frac{1}{2}gt^2\). Solving \(45 = \frac{1}{2} \times 10 \times t^2\) gives \(t^2 = 9\), so \(t = 3 \text{ s}\).
3. To find the distance fallen from 30 m s^-1 to 40 m s^-1, use \(v^2 = u^2 + 2gs\). Substituting \(40^2 = 30^2 + 2 \times 10 \times s\), we get \(1600 = 900 + 20s\). Solving gives \(s = 35 \text{ m}\).

(i) To find the maximum height of A, use the equation for vertical motion:

\(v = u - gt\)

where \(v = 0\) at maximum height, \(u = 5 \text{ m/s}\), and \(g = 9.8 \text{ m/s}^2\). Solving \(0 = 5 - 9.8t\) gives \(t = \frac{5}{9.8}\).

The height \(h\) is given by:

\(h = ut - \frac{1}{2}gt^2\)

Substitute \(t = \frac{5}{9.8}\) to find \(h_A\).

For B, use the same formula with \(u = 8 \text{ m/s}\) and the same \(t\). Calculate \(h_B\).

The difference in heights is \(h_B - h_A = 1.5 \text{ m}\).

(ii) To find when B is 0.9 m above A, solve:

\(h_B - h_A = 0.9\)

Using \(h = ut - \frac{1}{2}gt^2\) for both particles, set up the equation:

\(8t - \frac{1}{2}gt^2 - (5t - \frac{1}{2}gt^2) = 0.9\)

Simplify to \(3t = 0.9\), giving \(t = 0.3 \text{ s}\).

Substitute \(t = 0.3\) into \(h_A = 5t - \frac{1}{2}gt^2\) to find \(h_A = 1.05 \text{ m}\).

(a) To find the initial speed, use the equation of motion:

\(v^2 = u^2 + 2as\)

At the maximum height, \(v = 0\), \(a = -g\), and \(s = 45\) m. Thus,

\(0 = u^2 - 2g \times 45\)

Solving for \(u\), we get:

\(u^2 = 2g \times 45\)

\(u = \sqrt{2 \times 9.8 \times 45} = 30 \text{ m s}^{-1}\)

(b) To find the total time for which the speed is at least 10 m s^-1, use the equation:

\(v = u + at\)

Set \(v = 10 \text{ m s}^{-1}\), \(u = 30 \text{ m s}^{-1}\), and \(a = -g\):

\(10 = 30 - gt\)

\(gt = 20\)

\(t = \frac{20}{9.8} \approx 2 \text{ s}\)

The particle takes 2 s to reach 10 m s^-1 on the way up and another 2 s on the way down, so the total time is:

\(2 \times 2 = 4 \text{ s}\)

(a) To find the initial speed u, use the equation for maximum height:

\(0 = u^2 - 2 \times 10 \times 20\)

Solving for \(u\), we get:

\(u^2 = 400\)

\(u = 20 \text{ m/s}\)

(b) To find the total time for which P is at least 15 m above the ground, use the equation:

\(15 = 20t - \frac{1}{2} \times 10 \times t^2\)

Rearranging gives:

\(5t^2 - 20t + 15 = 0\)

Solving this quadratic equation, we find:

\(t = 1 \text{ or } t = 3\)

The total time is \(3 - 1 = 2 \text{ s}\).

(a) Using the equation of motion: \(s = ut + \frac{1}{2} a t^2\), where \(s = 24 \text{ m}\), \(t = 2 \text{ s}\), and \(a = -g = -9.8 \text{ m s}^{-2}\).

\(24 = u \times 2 - \frac{1}{2} \times 9.8 \times 2^2\)

\(24 = 2u - 19.6\)

\(2u = 43.6\)

\(u = 22 \text{ m s}^{-1}\)

(b) At maximum height, \(v = 0\), using \(v^2 = u^2 + 2as\):

\(0 = 22^2 - 2 \times 9.8 \times s\)

\(s = 24.2 \text{ m}\)

Height down in 1.8 s: \(s = \frac{1}{2} \times 9.8 \times 1.8^2 = 16.2 \text{ m}\)

\(h = 24.2 - 16.2 = 8 \text{ m}\)

Alternative method: Using \(v = u - gt\) with \(u = 22\) and \(v = 0\):

\(0 = 22 - 9.8t\)

\(t = 2.2 \text{ s}\)

\(h = 22 \times (2.2 - 1.8) - \frac{1}{2} \times 9.8 \times (2.2 - 1.8)^2\)

\(h = 8 \text{ m}\)

(a) To find the initial velocity v, we use the equation of motion:

\(v = u + at\)

where \(v = 0\) (velocity at the greatest height), \(a = -g = -10 \text{ m/s}^2\) (acceleration due to gravity), and \(t = 3 \text{ s}\).

Substituting the values, we get:

\(0 = v - 10 \times 3\)

\(v = 30 \text{ m/s}\)

(b) To find the greatest height, we use the equation:

\(v^2 = u^2 + 2as\)

where \(v = 0\), \(u = 30 \text{ m/s}\), and \(a = -10 \text{ m/s}^2\).

Substituting the values, we get:

\(0 = 30^2 + 2(-10)s\)

\(0 = 900 - 20s\)

\(20s = 900\)

\(s = 45 \text{ m}\)

Thus, the greatest height is 45 m.

(a) To find the height h of the building, we use the equation of motion for the first particle:

Using the equation for velocity, \(v = u + at\), with \(u = 40\), \(v = 0\), and \(a = -g\), we find the time to reach the highest point:

\(40 - gt = 0\) gives \(t = 4\) seconds.

The time to reach the top of the building is \(4 - \frac{1}{2} \times 10 \times 2^2 = 2\) seconds.

Using the equation for displacement, \(s = ut + \frac{1}{2} at^2\), with \(u = 40\), \(a = -g\), and \(t = 2\), we find:

\(h = 40 \times 2 - \frac{1}{2} \times 10 \times 2^2 = 60\) m.

(b) For the first particle, the height above ground is given by:

\(40t - \frac{1}{2} \times 10t^2\).

For the second particle, the height above the top of the building is:

\(20(t - 1) - \frac{1}{2} \times 10 \times (t - 1)^2\).

Setting the heights equal gives:

\(60 + 20(t - 1) - \frac{1}{2} \times 10 \times (t - 1)^2 = 40t - \frac{1}{2} \times 10t^2\).

Solving this equation, we find \(t = 3.5\) seconds.

(a) To find the greatest height above the ground reached by P, we use the equation for vertical motion:

\(v^2 = u^2 - 2gs\)

where \(v = 0\) m/s (at the highest point), \(u = 5\) m/s, and \(g = 9.8\) m/s².

\(0 = 5^2 - 2 \times 9.8 \times s\)

\(s = \frac{25}{19.6} = 1.25\) m

The height above the ground is \(2.8 + 1.25 = 4.05\) m.

(b) To find the length of time for which P is at a height of more than 3.6 m above the ground, we use the equation:

\(s = ut + \frac{1}{2}at^2\)

Let \(s = 3.6 - 2.8 = 0.8\) m, \(u = 5\) m/s, \(a = -9.8\) m/s².

\(0.8 = 5t - 4.9t^2\)

Solving the quadratic equation \(4.9t^2 - 5t + 0.8 = 0\), we find \(t = 0.2\) s and \(t = 0.8\) s.

The length of time for which P is above 3.6 m is \(0.8 - 0.2 = 0.6\) s.