Answer:

(a)

\(d=UT(\sin\alpha+\cos\alpha\tan\beta).\)

(b)

\(d=\frac{U^2}{g}.\)

(a) Let the initial speed of \(Q\) be \(V\). Since the particles collide, their horizontal displacements at time \(T\) are equal. Hence

\(V\cos\beta=U\cos\alpha.\)

The vertical displacement of \(P\) from its starting point after time \(T\) is

\(U\sin\alpha\,T-\frac12gT^2.\)

The particle \(Q\) starts a distance \(d\) above \(P\) and is projected below the horizontal, so its vertical position relative to the starting level of \(P\) after time \(T\) is

\(d-V\sin\beta\,T-\frac12gT^2.\)

At collision these are equal, so

\(U\sin\alpha\,T-\frac12gT^2 =d-V\sin\beta\,T-\frac12gT^2.\)

The gravitational terms cancel, giving

\(d=U\sin\alpha\,T+V\sin\beta\,T.\)

From \(V\cos\beta=U\cos\alpha\),

\(V=\frac{U\cos\alpha}{\cos\beta}.\)

Therefore

\(d=U\sin\alpha\,T+\frac{U\cos\alpha}{\cos\beta}\sin\beta\,T.\)

So

\(d=UT(\sin\alpha+\cos\alpha\tan\beta),\)

as required.

(b) If the particles collide when \(P\) is at its maximum height, the vertical velocity of \(P\) is zero at time \(T\). Thus

\(U\sin\alpha-gT=0,\)

so

\(T=\frac{U\sin\alpha}{g}.\)

Using \(\alpha=30^\circ\) and \(\beta=60^\circ\),

\(\sin30^\circ=\frac12, \qquad \cos30^\circ=\frac{\sqrt3}{2}, \qquad \tan60^\circ=\sqrt3.\)

Hence

\(\sin\alpha+\cos\alpha\tan\beta =\frac12+\frac{\sqrt3}{2}\cdot\sqrt3 =\frac12+\frac32 =2.\)

Also,

\(T=\frac{U/2}{g}=\frac{U}{2g}.\)

Therefore

\(d=UT(2)=2U\left(\frac{U}{2g}\right)=\frac{U^2}{g}.\)

Thus

\(d=\frac{U^2}{g}.\)