Question

A ball is thrown vertically upward with a velocity of 20m/s from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground (a)how high will the ball rise? (b) how long will it be before the ball hits the ground?
take g=10ms-2

Answer 1

H = u²/(2g)

H = 20²/(2×10)

H = 20 m

The ball will reach 20 m high from the point of projection.

It reaches 20 + 20.5 = 40.5m high from the ground.

t1 = 2u/g

t1 = 2×20/10

t1 = 4 seconds

If v is the velocity with which it hits the ground then,

v = √(u² + 2aS)

v = √(20² + 2×10×25)

v = 30 m/s

v = u + at

t2 = (v - u)/g

t2 = (30 - 20)/10

t2 = 1 second

Time taken for it to hit the ground = t1 + t2

= 4 + 1

= 5 seconds

Answer 2

Answer : -
V => 20m/s
H => 25 m
g => 10 Ms

t => 20÷10 => 2
t = 2s

Height => 1/2at^2
=> 1/2 × 20 × 4
=> 20 × 2
=> 40m

t => 2s
h => 40m

Hope this helps you