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?​

Answer 1

Answer:

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

Initial Velocity of 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

= 2 x 10 x 20 + 0

= 400

ˆ´ Final velocity of ball, v = 20 ms-1

t = (v-u)/a

ˆ´Time taken by the ball to strike = (20-0)/10

= 20/10

= 2 seconds

Hope it helps you

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Answer 2

Given:-

Height(h)=20m

Initial velocity(u)=0

Acceleration due to gravity(g)=10m/s²

To find:-

Final velocity(v)

Time taken(t)

Solution:-

By,h=1/2gt² we get-----

=>20=1/2×10×t²

=>20=5t²

=>t²=20/5

=>t²=4

=>t=2s

By,v=u+gt

=>v=0+10×2

=>v=20m/s

Thus,final velocity is 20m/s and time taken is 2s.