A ball is launched at 4.50 m/s at 60° above the horizontal. Find the maximum height and the time of flight of the ball.

Answer 1

V(vertical) = 4.5sin60 ≅ 3.9 m/s

V(horizontal) = 4.5 cos60 ≅ 2.3 m/s

When maximum height is reached, vertical velocity will drop to zero in time 't_max'.(symmetrically t_flight is twice as 't_max')

The formula for acceleration g = (V_f - V_i)/t

-9.8 = (0 - 3.9)/t_max (negative sign shows direction down )

Then, t_max = 9.8/3.9 = 2.5s

The time of flight is then, 5 seconds.

Maximum height can be found by the relation V_f² = V_i² + 2gh_max

0 = 3.9² - 2 x 9.8 x h_max

h_max = 0.78m