A ball is thrown vertically upwards with a velocity of 20ms^-1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0m from the ground. (a) How gh will the ball rise? (b) how long will it be before the ball ts the ground? Give me in detail, step by step please..
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(a) 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.
(b) 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
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.
(b) 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
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