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A ball is thrown upward. It leaves the hand with a velocity of 14.6 m/s, having been accelerated through a distance of 0.505 m. Compute the ball's upward acceleration, assuming it to be constant.

Respuesta :

Answer:

upward acceleration is 211.04 m/s²

Explanation:

given data

velocity = 14.6 m/s

distance = 0.505 m

to find out

upward acceleration

solution

we have given distance and velocity and ball is going upward

so acceleration will be calculated by  velocity formula that is

v² - u² = 2×a×s   ............1

here v is velocity and u is initial velocity and s is distance and a is acceleration

and u = o because starting velocity zero

put here all there value in equation 1

v² - u² = 2×a×s

14.6² - 0 = 2×a×0.505

solve it we get a

a = 211.04

so upward acceleration is 211.04 m/s²

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