Chapter 6, Problem 51A

a.

To determine

Time of dropping of box.

Answer to Problem 51A

  $t=14.29\text{s}$

Explanation of Solution

Given:

Height of plane, $y=1001\text{m}$

Speed od plane, $u_x=125\text{km/h}$

Formula Used:

The equation of motion is,

  $y=u_{y}t-\frac{1}{2}g{t}^2$

Here, $y$ is the displacement, $t$ is time and $g$ is acceleration due to gravity.

Calculation:

As $u_y=0$

So, from equation (1)

  $\begin{array}{l}t=\sqrt{\frac{-2y}{a}}\\ =\sqrt{\frac{\left(-2\right)\times \left(-1001\right)}{9.8}}\\ =14.29\text{s}\end{array}$

Conclusion:

The box should be dropped $t=14.29\text{s}$ before.

b.

To determine

Horizontal distance between plane and box.

Answer to Problem 51A

Horizontal distance, $x=496.5\text{m}$

Explanation of Solution

Given:

Height of plane, $y=1001\text{m}$

Speed od plane, $u_x=125\text{km/h}$

Formula Used:

The equation of motion is,

  $x=u_{x}t$

Here, $u$ is initial velocity, $t$ is time and $x$ is displacement.

Calculation:

As

  $\begin{array}{c}x=u_{x}t\\ =\left(125\times \frac{1000}{3600}\text{m/s}\right)\times 14.3\\ =496.5\text{m}\end{array}$

Conclusion:

Horizontal distance between the plane and the victims when the box is dropped is

  $x=496.5\text{m}$ .