Question

A ball is thrown upward with a speed of 38.0 m/s from the top of a building 240m tall.g) what will be the average speed of the ball from the time it is thrown until the time it strikes the ground?

Answer 1

The average speed formula is

`\bar{v}=\frac{total\text{ distance}}{total\text{ time elapsed}}`

We know that the total distance covered is 73.67 + 313.67 = 387.34 meters, this total distance includes the height of the building, and the distance covered by the ball on its way up and down (above the building).

Now, to find the total time elapsed, we have to find, first, the time taken to reach the highest point.

`v_f=v_0+gt`
`\begin{gathered} 0=38-9.8\cdot t \\ -38=-9.8t \\ t=(-38)/(-9.8) \\ t_(\max )\approx3.88\sec \end{gathered}`

Now, we have to find the time elapsed from the highest point to the ground. (h = 313.67m).

`\begin{gathered} h=v_0t+(1)/(2)gt^2 \\ 313.67=0+(1)/(2)\cdot9.8\cdot t^2 \\ 313.67=4.9t^2 \\ t=\sqrt[]{(313.67)/(4.9)} \\ t\approx8\sec \end{gathered}`

Now, we add.

`t_{\text{total}}=3.88\sec +8\sec =11.88\sec`

Once we have the total time elapsed, we can find the average speed of the ball.

`\begin{gathered} \bar{v}=(387.34m)/(11.88\sec ) \\ \bar{v}\approx32.6((m)/(s)) \end{gathered}`

Therefore, the average speed of the ball is 32.6 m/s.