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A ball is gently dropped from a height of 20m. 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 Text

Answer :
velocity `=20 m s^(-1)`; time =2 s
Solution :
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' <br> Initial Velocity of ball, u = 0 <br> Distance or height of fall, s = 20 m <br> Downward acceleration, a = 10 m `s^(-2)` <br> As we know, 2as = `v^(2)-u^(2)` <br> `v^(2)=2as+u^(2)` <br> = `2xx10xx20+0` <br> = 400 <br> `therefore` Final velocity of ball, v = 20 m`s^(-1)` <br> t = (v-u)/a <br> `therefore` Time taken by the ball to strike = (20-0)/10 <br> =20/10 <br> = 2 seconds

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