Suggested languages for you:

Americas

Europe

Q113P

Expert-verified
Found in: Page 39

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# 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?

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${}^{\mathbf{5}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{m}}$.

See the step by step solution

## 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)

$y={v}_{0}t+\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}$ … (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.

## 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:

$\mathbit{y}\mathbf{}\mathbf{=}{\mathbit{v}}_{\mathbf{0}}\mathbit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}\mathbit{g}{\mathbit{t}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{145}\mathbf{}\mathbit{m}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\left(9.8m/{s}^{2}\right){\mathbit{t}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbit{t}}^{\mathbf{2}\mathbf{}}\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{×}\mathbf{}\mathbf{145}}{\mathbf{9}\mathbf{.}\mathbf{8}}\mathbf{}{\mathbit{s}}^{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{44}\mathbf{}\mathbit{s}$

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:

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

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

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

$\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}\mathbf{}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}}{\mathbf{2}\left(-25g\right)}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\frac{{\left(0m/s\right)}^{\mathbf{2}}\mathbf{-}{\left(53.3m/s\right)}^{\mathbf{2}}}{\mathbf{2}\mathbf{-}\left(-25×9.8m/{s}^{2}\right)}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{m}$

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