Question

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Solve it

A ball is gently dropped from a height of 80m. If its velocity increases uniformly at the rate of 10ms² with what velocity will it stike the ground? After what time will it stirke the ground?

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Answer 1

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{40m/s,\:4sec}$

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Given & to find —

initial velocity of ball,u = 0

Final velocity of ball, v= ?

Distance through which the balls falls,s = 80 m

Acceleration, a = 10 m/s²

Time of fall, t =?

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We know

${v}^{2} - {u}^{2} = 2as$

$= > {v}^{2} - 0 = 2 \times 10 \times 80$

=> v = 40 m/s

Therefore velocity while strike the ground is 40 m/s

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Now for t we know

v = u + at

=> 40 = 0 + 10 t

=> t = 40/10

=> t = 4 sec

Therefore, time of fall is 4 sec

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Answer 2

Hii public

you are welcome in my ans

GIVEN

H = 80m

v = 0

dv/dt = -10m/s^2

V^2 = U^2 +2ah

0 = U^2 - 2*10*80

u^2 = 1600

u =40m/s

v = u + at

40 = 10t

t = 4s

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Hope it may helps you