Question

A juggler throws a bowling pin straight up with an initialspeed of 8.20 m/s. How much time elapses until the bowling pinreturns to the juggler’s hand?

Short answer

The total time elapsed by the bowling pin is 1.67 s.

Step-by-step solution

Given data  

Total displacement is$\Delta y=0\text{m}$

The initial velocity is$u_{oy}=8.2m/s$ and acceleration is given as$a_y=-9.8m/s^2$

Solution

Using Newton’s second law of motion,

$\Delta y=u_{oy}t+\frac{1}{2}{a}_y{t}^2$

Substituting the values in the above expression will give,

$\begin{array}{rcl}0& =& 8.2m/s\times t+\frac{1}{2}\left(-9.8m/s^2\right)\times t^2\\ t& =& \frac{8.2m/s}{4.9m/s^2}\\ t& =& 1.67s\end{array}$

Hence, the time elapsedis$t=1.67s$