A ball is gently dropped from a height of 20 metres if its velocity increase uniformly at rate of 10 M/s2 square what will be the velocity when it he when it hit the ground
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by using Newton's third eq. of motion
v^2 –u^2 = 2as
v^2 -0^2 = 2×10×20
v^2 = 20^2
=> v = √(20^2)
v = 20 m/s.
v^2 –u^2 = 2as
v^2 -0^2 = 2×10×20
v^2 = 20^2
=> v = √(20^2)
v = 20 m/s.
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Hello
Height from where the ball is dropped is 20m
Rate of change of velocity is acceleration
So acceleration of the ball is 10m/s²
Initially it had no velocity so u=0
Using the newtons law of motion
v²-u²=2as
Where v is e final velocity
v²-0²=2*10*20
v²=20*20
v=√(20*20)
v=20
So the final velocity i.e the velocity of the ball when it hits the ground is 20m/s
Height from where the ball is dropped is 20m
Rate of change of velocity is acceleration
So acceleration of the ball is 10m/s²
Initially it had no velocity so u=0
Using the newtons law of motion
v²-u²=2as
Where v is e final velocity
v²-0²=2*10*20
v²=20*20
v=√(20*20)
v=20
So the final velocity i.e the velocity of the ball when it hits the ground is 20m/s
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