Chapter 3, Problem 89A

(a)

To determine

To Compute: Distance covered by the pack fall.

Answer to Problem 89A

  $s=275.6\text{ m}$

Explanation of Solution

Given:

Initial speed of pack, $u=0$

Final speed, $v=73.5\text{ m/s}$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Formula used:

Third equation of kinematics:

  $v^2=u^2+2as\mathrm{..........}\text{(i)}$

Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

Calculation:

Let height be $s$ .

a = acceleration due to gravity = g

From equation (i):

  $\begin{array}{c}s=\frac{v^2-u^2}{2g}\\ =\frac{{\left(73.5\right)}^2-0^2}{2\times 9.8}\\ =275.6\text{ m}\end{array}$

Conclusion:

Distance covered by the pack: $275.6\text{ m}$

(b)

To determine

To Compute: Time taken by the pack to fall.

Answer to Problem 89A

  $t=7.5\text{ s}$

Explanation of Solution

Given:

Initial speed of pack, $u=0$

Final speed, $v=73.5\text{ m/s}$

Acceleration due to gravity, $g=9.8{\text{ m/s}}^2$

Formula used:

First equation of kinematics:

  $v=u+at$

Where, v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Calculation:

Let the time taken be $t$ .

Plugging in the given values in the formula:

  $\begin{array}{c}73.5=0+9.8\times t\\ t=7.5\text{ s}\end{array}$

Conclusion:

Time taken by the pack to fall: $7.5\text{ s}$