A ball is thrown upward with a velocity of Vi. The ball reaches a height of 39 meters. Determine the initial velocity of the ball and the time of the entire trip.

27.6 m/s

5.64 s

Explanation:

Given:

Δy = 39 m

a = - 9.8 m/s²

v = 0 m/s

Find: v₀

v² = v₀² + 2aΔy

(0 m/s) ² = v₀² + 2 (-9.8 m/s²) (39 m)

v₀ = 27.6 m/s

Find: 2t

Δy = vt - ½ at²

(39 m) = (0 m/s) t - ½ (-9.8 m/s²) t²

t = 2.82 s

2t = 5.64 s