Question

A stone is projected vertically upwards from the ground with a velocity of 49 ms^-1. At the same time, another stone is dropped from a height 98m to fall freely along the same path as the first. Find where and when the two stones meet each other. 

Answer 1

Let P be the stone dropped from B at a height 98m from the ground A. Q is the other stone thrown up vertically upwards with a velocity 49ms^-1 at the same instant. Let the two stones meet at C after a time t seconds.

For the stone P, we have

u = 0, a = + g, s = + h (=BC), t = t

From s = ut + 1/2 at², we have,

+h = 0 × t + 1/2 gt² or h = 1/2 gt² ………. (1)

For the stone Q,

u = 49ms^-1,

a = -g,

s = +AC = +(98 – h),

(98 – h) = 49t – 1/2 gt²

Using eqn. (1) in (2), we have,

(98 – h) = 49t – h or 49t = 98

∴ t = 2s.

Using this in (1), we have

h = BC = 1/2 × 9.8 × 4 = 19.6 m

∴ AC = 96 – 19.6 = 76.4 m

∴ Two stones meet at a height of 76.4 m from the ground, after 2 seconds.