a ball is thrown straight upward with an initial velocity of 16 m/s. use g+/- 10 m/s^2 for computations. how much time is required for the ball to reach its initial height

t = 1.6s

Explanation: Use Fab 5 equation to solve for t (d is irrelevant so use equation without d)
vi = 16m/s, a = -10m/s^2, vf = 0m/s, t = ?

(Assume up is positive so acceleration is negative!!!)

a = vf - vi / t
-10 = 0 - 16 / t
-10t = -16
t = 1.6s

Hope this helps! :)

Explanation:

At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is