Question

At the base of a vertical cliff, a model rocket, starting from rest, is launched upwards at t = 0 with a time-varying acceleration given by

ay(t) = A − Bt

where A and B are positive constants. Also at t = 0, a small stone is released from rest from the top of the cliff at a height h directly above the rocket. (This height h is higher than the maximum height reached by the rocket.) The stone hits the rocket at the instant when the rocket reaches its maximum height. The gravitational acceleration of magnitude g is downward. You may neglect air resistance. Determine an expression for the initial height h from which the stone was dropped in terms of the constants A, B, and g. 

Answer 1

$a(t)=A-Bt$

$V=\int adt=At- \frac{Bt^2}{2} \implies v=0 \space at \space t=\frac{2A}{B}$

$x-\int Vdt=\frac{At^2}{2}- \frac{Bt^3}{6} =t^2(\frac{A}{2}- \frac{Bt}{6})$

Distance travelled by stone in time t, $D=0.5gt^2$

$D=0.5\times g \times \frac{4A^2}{B^2}=\frac{2A^2}{B^2}g$

But we know x+D=h

$h=\frac{A^2}{B}(2g+\frac{4A}{6})$

$h=\frac{A^2}{B^2}(2g+\frac{2A}{3})$

$h=\frac{2A^2(3g+A)}{3B^2}$