Question

A ball is gently dropped from height of 20 m. If its velocity increases uniformly at the rate of 10ms^-2, with what velocity will it strike the ground? After what time will it strike the ground?​

Answer 1

Answer:

Let us assume, the final velocity with which the ball will strike the ground be ‘v’ and the time it takes to strike the ground be ‘t’.

Given parameters

Initial Velocity of the ball (u) = 0

Distance or height of fall (s) = 20 m

Downward acceleration (a) = 10 m s-2

As we know

2as = v2 – u2

v2 = 2as +  u2

v2 = (2 x 10 x 20 ) + 0

v2 =  400

Final velocity of ball (v) = 20 ms-1

t = (v – u)/a

Time taken by the ball to strike (t) = (20 – 0)/10

t = 20/10

t = 2 seconds

The final velocity with which the ball will strike the ground is (v) =  20 ms-1

The time it takes to strike the ground (t) = 2 seconds

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Answer 2

Explanation:

velocity = displacement/time taken

time = displacement / velocity

= 20 / 10

= 2s