Q37E - 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$∆y=0\mathrm{m}$.

The initial velocity is$u_{ay}=8.2 \mathrm{m}/\mathrm{s}$ and acceleration is given as${\mathrm{a}}_{\mathrm{y}}=-9.8 \mathrm{m}/{\mathrm{s}}^2$

Step 2: Solution

Using Newton’s second law of motion,

$∆y=u_{ay}t+\frac{1}{2}{a}_y{t}^2$

Substituting the values in the above expression will give,

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

Hence, the time elapsed is$t=1.67 \mathrm{s}$