Question

A ball is allowed to fall from the top of a tower 200 m high. At the same instant, another ball is thrown
vertically upwards from the bottom of the tower with a velocity of 40 m/s. When and where the two balls meet?​

Answer 1

Answer:

Let us assume that the two balls meet at a height 'h' after time 't' above the ground.

For the ball dropped from the top of the tower:

Distance covered by the ball is (200 - h) m

Here u = 0 ; s = (200 - h) m and g = 9.8 ms-2

→ s = ut + 1/2 at2

or, 200 - h = 0 × t + 1/2 at2

or, 200 - h = 4.9t2 ............(Equation 1)

For the ball thrown vertically upwards:

u = 40ms-1 ; s = h and g = -9.8ms-2 (-ve value of g since thrown upwards)

s = ut + 1/2 at2

h = 40 × t + 1/2(-9.8)t2

or, h = 40t - 4.9t2 ............(Equation 2)

Adding equations (1) and (2), we get

200 - h + h = 4.9t2 + 40t - 4.9t2

On solving we get

t = 5 s

Substituting t = 5 seconds in equation 1, we get

200 - h = 4.9 × (5)2

or, h = 200 - 4.9 × 25

= 200 - 122.5

= 77.5 m

Thus, the two balls meet at a height 77.5 m from the ground after 5 s.