Question

All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will take for it to reach its peak (assume no air resistance). (Answer to two decimal places).

Answer 1

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

$v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}$

since $v_y = 0$ at the top of its trajectory. Plugging in the numbers,

$t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}$