Q37Q - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Given data

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

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

Step 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,

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

Hence, the time elapsedis $t = 1.67s$