Question

You shoot an arrow into the air. Two seconds later (2.00 s ) the arrow has gone straight upward to a height of 29.0 m above its launch point.

Answer 1

t = 5.91 seconds

Step-by-step explanation:

The time it takes for the arrow to fall back to its origin can be found by algebraic manipulation of the kinematic equation:

`d = (1)/(2) *a*t^2\\t^2 = (2d)/(a) \\t =\sqrt{(2d)/(a)}\\t=\sqrt{(2(29m))/(9.8m/s^2)} = \sqrt{(58)/(9.8)}\\t= 5.91 s`