Question

From the top of the tower 45 m high, two stones are released. One vertically downwards and the other with a horizontal velocity of 30 m/s. How long will each stone to strike the ground and how far from the tower will each stone strike the ground

Answer 1

Initial downward velocity of both the stones is 0.
That means, both will reach the ground in same time.

s = 45 m, u = 0, g = 10 m/s ^2

By second equation of motion,
s =ut + (1/2) gt^2=> 45 = 0 + (1/2) x 10 x t^2=> 45 = 5 t^2=> t^2 = 9 s
=> t = 3 s

Hence, both stones will reach the ground in 3 s.

We know that the stone which is projected horizontally has a horizontal velocity of 30 m/s.

So its distance from the ground = velocity x time                                                   = 30 x 3
                                                   = 90 m
It will fall there on the ground.

The other stone will fall just by the tower.

Answer 2

Initial downwards velocity of both the stones is a that means both will reach the ground in sometime

s= ut +(1/2)g 1^2=45=0+(1/2)×10

×t^2 =45=5t^2 =t^2=05

We know that the stone which is project.

horizontal velocity is 30m/s so it's distance from the ground

The other stone will fall just by tower.

Thank You