Feb/Mar 2023 p42 q2
3860
A particle P is projected vertically upwards from horizontal ground with speed 15 m s-1.
(a) Find the speed of P when it is 10 m above the ground.
At the same instant that P is projected, a second particle Q is dropped from a height of 18 m above the ground in the same vertical line as P.
(b) Find the height above the ground at which the two particles collide.
Solution
(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 \(s_P\):
\(s_P = 15 \times \frac{15}{g} - \frac{1}{2}g \left(\frac{15}{g}\right)^2\)
\(s_P = 10.8 \text{ m}\)
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