Chapter 3, Problem 122P

To determine

The time at which the packet fell off the back ramp.

Explanation of Solution

Given: The speed of the airplane is $275\text{mi}/\text{hr}$ .

The angle is $37°$ above the horizontal.

The distance is $7.5\text{km}$ .

Formula used: Draw the diagram for the representation of the package when it fell off the ramp.

  

Write the expression for the distance in x direction.

  $x_d=v_{x,d}{t}_d$

Here, $x_d$ is the distance in the x direction, $v_{x,d}$ is the velocity in the x direction and $t_d$ is the time.

Write the expression for the distance in y direction.

  $y_d=v_{y,d}{t}_d$

Here, $y_d$ is the distance in the y direction, $v_{y,d}$ is the velocity in the y direction and $t_d$ is the time.

Write the expression for the x coordinate along the parabolic trajectory of the airplane.

  $x(t)=x_d+v_{x,d}(t-t_d)$   ...... (1)

Here, $x(t)$ is the position as the function of time, $v_{x,d}$ is the velocity and $t_d$ is the time.

Write the expression for the y coordinate along the parabolic trajectory of the airplane.

  $y(t)=y_d+v_{y,d}(t-t_d)-\frac{1}{2}g{(t-t_d)}^2$   ...... (2)

Here, $y(t)$ is the position as the function of time, $g$ is acceleration due to gravity.

Substitute $v_{x,d}{t}_d$ for $x_d$ in equation (1).

  $x(t)=v_{x,d}{t}_d+v_{x,d}(t-t_d)$

Substitute $v_{y,d}{t}_d$ for $y_d$ in equation (2).

  $y(t)=v_{y,d}{t}_d+v_{y,d}(t-t_d)-\frac{1}{2}g{(t-t_d)}^2$

Simplify the above equation

  $y(t)=v_{y,d}{t}_d-\frac{1}{2}g{(t-t_d)}^2$

Write the expression for the velocity component with respect to x axis.

  $v_{x,d}=v_p\cos {\theta }_0$   ...... (3)

Here, $v_p$ is the velocity of the plane and ${\theta }_0$ is the angle.

Write the expression for the velocity component with respect to y axis.

  $v_{y,d}=v_p\sin {\theta }_0$   ...... (4)

At impact, the horizontal distance is $x_l$ and vertical distance is zero.

Write the expression for the distance to impact.

  $x_l=v_{x,d}{t}_t$

Simplify the above equation for $t_t$ .

  $t_t=\frac{x_l}{v_{x,d}}$

Substitute $\frac{x_l}{v_{x,d}}$ for $t_t$ and $0$ for $y(t)$ in equation (2).

  $0=v_{y,d}{t}_d-\frac{1}{2}g{\left(\frac{x_l}{v_{x,d}}-t_d\right)}^2$

Substitute $\frac{x_l}{v_{x,d}}$ for $t_d$ in the above equation.

  $0=v_{y,d}\frac{x_l}{v_{x,d}}-\frac{1}{2}g{\left(\frac{x_l}{v_{x,d}}-t_d\right)}^2$

Solve the above equation using binomial expansion.

  $t_d^2-\frac{2x_l}{v_{x,d}}{t}_d+\left(\frac{x_l^2}{v_{x,d}^2}-\frac{2x_l{v}_{y,d}}{g{v}_{x,d}}\right)=0$   ...... (5)

Calculation:

Substitute $275\text{mi}/\text{hr}$ for $v_p$ and $37°$ for ${\theta }_0$ in equation (3).

  $\begin{array}{l}{v}_{x,d}=\left(\text{275}\frac{\text{mi}}{\text{hr}}\text{×1609}\frac{\text{m}}{\text{mi}}\text{×}\frac{\text{1}\text{hr}}{\text{3600}\text{s}}\right)\cos 37°\\ v_{x,d}=98.16\text{m}/\text{s}\end{array}$

Substitute $275\text{mi}/\text{hr}$ for $v_p$ and $37°$ for ${\theta }_0$ in equation (4).

  $\begin{array}{l}{v}_{y,d}=\left(\text{275}\frac{\text{mi}}{\text{hr}}\text{×1609}\frac{\text{m}}{\text{mi}}\text{×}\frac{\text{1}\text{hr}}{\text{3600}\text{s}}\right)\sin 37°\\ v_{y,d}=73.97\text{m}/\text{s}\end{array}$

Substitute $7.5\text{km}$ for $x_l$ , $73.97\text{m}/\text{s}$ for $v_{y,d}$ , $98.16\text{m}/\text{s}$ for $v_{x,d}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (5).

  $\begin{array}{l}{t}_d^2-\frac{2\left(7.5\text{km}\right)}{98.16\text{m}/\text{s}}{t}_d+\left(\frac{{\left(7.5\text{km}\right)}^2}{{\left(98.16\text{m}/\text{s}\right)}^2}-\frac{2\left(7.5\text{km}\right)73.97\text{m}/\text{s}}{\left(98.16\text{m}/\text{s}\right)}\right)=0\\ t_d^2-\left(152.8\text{s}\right)t_d+4686s^2\\ t_d=42\text{s}\end{array}$

Conclusion:

Thus, the time at which the packet fell off the back ramp is $42\text{s}$ .