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.
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
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
1/2 m v^2 = m g h ... h = v^2 / (2 g)
0 = -4.9 t^2 + 14 t
po
what is the expression for time?
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
so h = 0 at t = 0 and at t = 14 / 4.9