Question

An arrow is shot straight up at an initial velocity of 200 m/s. How long will it be in the air before beginning to fall?

Answer 1

`20.4 sec`

Explanation:

Using Calculus to derive the equations:
position equation: `s(t) = 1/2g t^2 + v_ot+h_o`
where `v_o` = initial velocity and `h_o` = initial height

velocity equation : `s'(t) = v(t) = g t + v_o`

Due to gravity: `g = -9.8 m/s^2` the arrow at its maximum helight will have its `v = 0`

Using the velocity equation: `0 = -9.8 t + 200`

Solving for `t: 9.8t = 200` or `t = 200/9.8 ~~ 20.4` sec