Question

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

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Answer 1

Answer:

Height, s=20m

Acceleartion , a=10m/s

2

By 3rd equation of motion

v

2

=u

2

+2as

v

2

=0+2×10×20=400

⇒v=20m/s

So, we conclude final velocity , v=20m/s

again, by 1st equation of motion , v=u+at

⇒t=

a

v−u

⇒t=

10

20

=2s

So, it will strike the ground after 2s

Answer 2

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

Explanation:

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