Question

1. A stone is thrown vertically upward with velocity of 15 m/s at the same time, 10 m vertically above a second stone is allowed to fall. After what time and at what height do they collide (take g = 10 ms^-2)​

Answer 1

Let the ball at the ground be A and that at the top be B.

Assume that you see the motion of A while sitting at B.

You(B) are obviously at rest w.r.t. yourselves. However you(B) and A have the same acceleration in the same direction. Thus acceleration of A w.r.t. you(B) is 0.

Thus, to you, it will appear as if A is travelling towards you with an uniform speed of 20m/s.

Dividing the distance 20m with that speed, we get that A reaches you(B) in 1s.

Now, you have considered these motions and time by assuming you were stationed at B. However, the time taken for A to meet B shouldn't be dependent on which reference frame you assume. Thus, time taken for A and B to collide is 1s.

Since B falls freely, he covers a distance of  1/2(g)(1²)=5  (assuming g = 10m/s²)

Thus, they meet at a height of (20–5)m = 15m from the ground.