Q72P - University Physics with Modern Physics

Question

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 m away at an average speed of 3.00 m/s, returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?

Step-by-Step Solution

Answer

(a) Minimum initial speed is 17.96 m/s.

(b) Initial position is 16.44 m.

Step 1: identification of given data

Average speed v=3.00m/s

Distance d=5.50m

Step 1: Calculation of speed

As we know that

 $$\begin{gathered} \mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}} \\ \mathrm{t}=\frac{5.5\times 2}{3} \\ \mathrm{t}=3.67\mathrm{s} \\ \mathrm{Also} \\ {\mathrm{v}}_0\mathrm{t}=\frac{1}{2}{\mathrm{gt}}^2 \\ {\mathrm{v}}_0=\frac{1}{2}\times 9.8\times 3.67 \\ {\mathrm{v}}_0=17.96\mathrm{m}/\mathrm{s} \end{gathered}$$

Step 2: Calculation of position

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}} \\ \mathrm{t}=\frac{5.5}{3} \\ \mathrm{t}=1.83\mathrm{s} \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that} \\ \mathrm{y}={\mathrm{v}}_0\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{y}=(17.96\times 1.83)-\frac{1}{2}\times 9.8\times 1.{83}^2 \\ \mathrm{y}=16.44\mathrm{m} \end{gathered}$$

It is 16.44 m high above its initial position when ball just as she reaches the table