Answer:

(a) \(A\left(\frac{105}{2},\frac{35}{4}\right)\).

(b) \(15\sqrt2\text{ m s}^{-1}\).

(c) \(\frac52\text{ m}\).

Since \(\tan\theta=\frac43\),

\(\sin\theta=\frac45, \qquad \cos\theta=\frac35.\)

The initial horizontal and vertical components of velocity are therefore

\(25\cos\theta=15, \qquad 25\sin\theta=20.\)

(a) At time \(t\), the velocity components are

\(v_x=15, \qquad v_y=20-10t.\)

At \(A\), the direction of motion makes an angle of \(45^\circ\) with the downward vertical. Hence the velocity is at \(45^\circ\) below the horizontal, so

\(\frac{v_y}{v_x}=-1.\)

Thus

\(\frac{20-10t}{15}=-1.\)

So

\(20-10t=-15, \qquad 10t=35, \qquad t=\frac72.\)

The coordinates of \(A\) are

\(x=15\left(\frac72\right)=\frac{105}{2},\)

and

\(y=20\left(\frac72\right)-\frac12(10)\left(\frac72\right)^2 =70-\frac{245}{4} =\frac{35}{4}.\)

Therefore

\(A\left(\frac{105}{2},\frac{35}{4}\right).\)

(b) Immediately before the collision at \(A\), the velocity components are

\(v_x=15, \qquad v_y=-15.\)

So the speed is

\(\sqrt{15^2+(-15)^2}=15\sqrt2.\)

Thus the speed immediately before collision is

\(15\sqrt2\text{ m s}^{-1}.\)

(c) The barrier is smooth and is inclined at \(45^\circ\) to the horizontal, so the incoming velocity at \(A\) is normal to the barrier. The tangential component is zero, and the speed after rebounding is therefore

\(w=e(15\sqrt2)=\frac19(15\sqrt2)=\frac53\sqrt2.\)

After the collision, the particle is projected from height \(\frac{35}{4}\) at an angle of \(45^\circ\) to the horizontal. Let \(X\) be the horizontal distance travelled after the collision before reaching the ground.

Using the trajectory equation from the point \(A\), the vertical displacement to the ground is \(-\frac{35}{4}\), so

\(-\frac{35}{4}=X\tan45^\circ -\frac{10X^2}{2w^2\cos^245^\circ}.\)

Since \(\tan45^\circ=1\), \(\cos^245^\circ=\frac12\), and \(w^2=\frac{50}{9}\),

\(-\frac{35}{4}=X-\frac{10X^2}{\frac{50}{9}}.\)

Hence

\(-\frac{35}{4}=X-\frac95X^2.\)

So

\(\frac95X^2-X-\frac{35}{4}=0.\)

The positive root is

\(X=\frac52.\)

Therefore the horizontal distance travelled after striking the barrier is

\(\frac52\text{ m}.\)