A projectile is fired at 65.0° above the horizontal. Its initial speed is equal to 37.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored.At what time after being fired does the projectile reach this maximum height?

Answer 1

Vo = 37.5m/s @ 65o.
Yo = 37.5*sin65 = 34 m/s.

Y = Yo + gt.
t = (Y-Yo)/g = (0-34)/-9.8 = 3.47 s.