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 upward from the ground with a velocity of 25m/s.calculate when and where the two stone will meet.

Answer 1

heya...

Consider, a stone fall from the top of a tower (h) and distance covered by a stone at time  t

Height (h) = 100 m

So,  x - x0 = u0t + 1/2 gt2

Initial velocity  = 0

or, 100 - x = 0 +1/2 gt2 .....eq (1)

Distance covered by stone from ground

Where initial velocity (u0) = 25 m/s

x = u0t  - 1/2 gt2

or, x =( 25 X t) - 1/2 gt2  .....eq (2)

Combine .....eq (1) and .....eq (2)

So, 100 = 25t

t = 4

therefore, x = 25X4 -1/2 X9.8X 42 = 100 - 78.4 = 21.6

hope helped..