Asked by po

An archer fires an arrow straight up into the air with a speed vo = 14 m/s. Neglect air resistance.
Find the maximum height h reached by the arrow, in meters.
Write an expression for the time the arrow is in the air until it returns to launch height in terms of known quantities.

Answers

Answered by R_scott
K.E. at launch equals P.E. at peak

1/2 m v^2 = m g h ... h = v^2 / (2 g)

0 = -4.9 t^2 + 14 t
Answered by po
what is the expression for time?
Answered by Anonymous
0 = -4.9 t^2 + 14 t = t (14 - 4.9 t)
so h = 0 at t = 0 and at t = 14 / 4.9
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