Question

The Zero Gravity Research Facility at the NASA Glenn Research Center includes a ${}^{\mathbf{145}\mathbf{}\mathbf{m}}$drop tower. This is an evacuated vertical tower through which, among other possibilities, a ${}^{\mathbf{1}\mathbf{}\mathbf{m}}$diameter sphere containing an experimental package can be dropped. (a) How long is the sphere in free fall? (b) What is its speed just as it reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average deceleration of ${}^{\mathbf{25}\mathbf{}\mathbf{g}}$as its speed is reduced to zero. Through what distance does it travel during the deceleration?

Short answer

1. The time spent by sphere in a free fall is${}^{\mathbf{5}\mathbf{.}\mathbf{44}\mathbf{}\mathbf{s}}$.

2. The speed of the sphere as it reaches a catching device at the bottom of the tower is ${}^{\mathbf{53}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$.

3. The distance travelled by the sphere during deceleration is${}^{\mathbf{5}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{m}}$.

Step-by-step solution

Given data

The height of the tower, ${}^{\mathbf{y}\mathbf{=}\mathbf{}\mathbf{145}\mathbf{}\mathbf{m}}$

Understanding the free fall acceleration

In free fall acceleration, objects accelerates vertically downward at constant rate. This constant acceleration is represented by g and it is known as free fall acceleration.

The kinematic equations under free falls is written as:

$v=v_0+gt$ … (i)

$y=v_{0}t+\frac{1}{2}g{t}^2$ … (ii)

$v^2={v_0}^2+2gy$ … (iii)

Here, ${}^{{\mathbf{v}}_{\mathbf{0}}}$is the initial velocity, ${}^{\mathbf{v}}$is the final velocity, ${}^{\mathbf{t}}$ is the time, ${}^{\mathbf{g}}$is the acceleration and ${}^{\mathbf{y}}$is the vertical displacement.

(a) Determination of the time spent by sphere in a free fall

Using equation (ii), the time spent by sphere in a free fall is calculated as follows:

$$\begin{gathered} \mathit{y}\mathbf{}\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}\mathit{g}{\mathit{t}}^{\mathbf{2}} \\ \mathbf{145}\mathbf{}\mathit{m}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\left(9.8m/s^2\right){\mathit{t}}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{t}}^{\mathbf{2}\mathbf{}}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{145}}{\mathbf{9}\mathbf{.}\mathbf{8}}\mathbf{}{\mathit{s}}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{44}\mathbf{}\mathit{s} \end{gathered}$$

Therefore, the sphere is in free fall for${}^{\mathbf{5}\mathbf{.}\mathbf{44}\mathbf{}\mathbf{s}}$.

(b) Determination of the speed of the sphere as it reaches a catching device

Using equation (iii), the speed of the sphere is calculated as follows:

$$\begin{gathered} {\mathbf{v}}^{\mathbf{2}\mathbf{}}\mathbf{=}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{2}\mathbf{}\mathbf{gy} \\ {\mathbf{v}}^{\mathbf{2}\mathbf{}\mathbf{=}}\mathbf{0}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\left(9.8m/s^2\right)\left(145m\right) \\ \mathbf{v}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{2842}\mathbf{}}\mathbf{m}\mathbf{/}\mathbf{s} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{53}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$

Therefore, the speed of the sphere as it reaches a catching device at the bottom of the tower is ${}^{\mathbf{53}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$.

(c) Determination of the distance traveled during the deceleration

For decelerated motion, ${}^{\mathbf{v}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}}$, ${\mathit{v}}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{53}\mathbf{.}\mathbf{3}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$and ${}^{\mathbf{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{25}\mathbf{}\mathbf{g}}$

Using equation (ii), the distance traveled during deceleration is calculated as follows:

$$\begin{gathered} \mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}}{\mathbf{2}\left(-25g\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{{\left(0m/s\right)}^{\mathbf{2}}\mathbf{-}{\left(53.3m/s\right)}^{\mathbf{2}}}{\mathbf{2}\mathbf{-}\left(-25\times 9.8m/s^2\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{m} \end{gathered}$$

Therefore, sphere travels ${}^{\mathbf{5}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{m}}\mathbf{}$during deceleration.