Question

A ball is thrown vertically upward with a speed of 15 m/s. Determine the time of flight when it

returns to its original position.

Answer 1

Explanations & Calculations

• Since it returns to the original position, the displacement is zero. Therefore, using the equation$\small S=\small ut+\frac{1}{2}at^2$ for the ball's motion airborne time can be found.

$\qquad\qquad \begin{aligned} \small 0&=\small 15\times t+\frac{1}{2}\times(-g)\times t^2 \\ &=\small 15t-4.9t^2 \\ \small 0&=\small t\Big(15-4.9t\Big)\\ \\ \small t &=\begin{cases} \small 0:extraneouse\\ \frac{15}{4.9}=\small 3.1: accepted \end{cases} \end{aligned}$

• Therefore, the time of flight is $\small \bold{3.1\,s}$