Question

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.

Answer 1

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

Answer 2

what is the expression for time?

Answer 3

0 = -4.9 t^2 + 14 t = t (14 - 4.9 t)
so h = 0 at t = 0 and at t = 14 / 4.9