Answer to Question #101782 in Physics for kwikiriza steven

Question

A stone is thrown vertically upwards from a tower at 29.4m/s, after 4 sec the other is released to fall freely. Prove that the first stone will overtake the second just 4 sec after releasing the second stone

Answer 1

$$h_1=h+v(t+T)-0.5g(t+T)^2$$

$$h_2=h-0.5gT^2$$

$$h_1=h_2\to 2v(t+T)=g(t^2+2tT)$$

$$2(29.4)(4+T)=9.8(4^2+2(4)T)$$

$$T=4\ s$$

The first stone will overtake the second just 4 sec after releasing the second stone.