Question

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Answer 1

Here, h = 100 m

Let the two stones meet after t seconds at point P, which is at a height x above the ground, as shown in the figure.

For stone 1,

u = 0, h = (100 - x) m,

a = g = 9.8 m/s²

From s = ${\displaystyle ut+\frac{1}{2} a{t}^2}$

(100 - x) = ${\displaystyle 0+\frac{1}{2}\times 9.8t^2}$

= 4.9t²         ...(i)

For stone 2,

u = 25 m/s, h = x,

a = - g = - 9.8 m/s²

From s = ${\displaystyle ut+\frac{1}{2} a{t}^2}$

x = ${\displaystyle 25t+\frac{1}{2}(-9.8)t^2}$

= 25t - 4.9t²        ...(ii)

Adding equations (i) and (ii)

100 - x + x = 25t

⇒ t = ${\displaystyle \frac{100}{25}}$

= 4 s

From equation (i),

100 - x = 4.9 × (4)²

100 - x = 78.4

x = 100 - 78.4

x = 21.6 m