Asked by meghan

a student drops a ball from a window 3.5m above the sidewalk. the ball accelerates at 9.80m/s^2. How fast is it moving when it hits the sidewalk.

Answers

Answered by Henry
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*3.5 = 68.6,
Vf = 8.28m/s.
Answered by A
Vf^2= Vi^2 + 2ad
Vf^2= 2(-9.80)(3.5)
Vf^2= -68.6
(square root Vf^2 and drop the - on 68.6 and square root it)
Vf= 8.3 m/s down
(answer is rounded up to 2 sig figs)
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