Q113P - Fundamentals Of Physics

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?

Step-by-Step Solution

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 1: Given data

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

Step 2: 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)

$v^{2 }={v_0}^{2 }+2 gy$                                                                                       … (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.

Step 3: (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: 

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

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

Step 4: (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: 

 

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

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

Step 5: (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: 

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

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