Chapter 3, Problem 35P

(a)

To determine

The distance at which goods must be dropped.

Answer to Problem 35P

The horizontal distance travelled by the plane is 480.6m.

Explanation of Solution

Given:

Plane is 235 m above the ground level and moving with speed of 69.4 m/s.

Initial speed, u = 69.4 m/s

Height, h = 235 m

Calculation:

The expression for the velocity of the plane in the horizontal is given as:

  $\begin{array}{l}{v}_x=\frac{x}{t}\\ \text{where,}\\ v_x=\text{Horizontal velocity}\\ x=\text{Horizontal distance}\\ t=\text{Time}\end{array}$

The vertical velocity of the plane will be 0 because the plane is moving in the horizontal direction.

  $v_y=0$

As the motion of the plane is projectile motion hence, the kinematic equation for the plane is given as:

  $\begin{array}{l}h=v_{y}t+\frac{1}{2}a{t}^2\\ \text{where,}\\ h=\text{height of the plane from the ground}\\ a=\text{acceleration}\end{array}$

Here, the acceleration of the plane is equal to the acceleration due to the gravity because it will travels under gravity.

  $a=g$

Substituting the known values into the above equation:

  $\begin{array}{l}h=0\left(\frac{x}{v_x}\right)+\frac{1}{2}\left(g\right){\left(\frac{x}{v_x}\right)}^2\\ h=\frac{1}{2}g{\left(\frac{x}{v_x}\right)}^2\\ x=v_x\sqrt{\frac{2h}{g}}\end{array}$

Put the known values into the above equation:

  $\begin{array}{l}x=\left(69.4\text{m/s}\right)\left(\sqrt{\frac{2\left(235\right)}{9.8{\text{m/s}}^{\text{2}}}}\right)\\ x=480.6\text{m}\end{array}$

Hence, the horizontal distance travelled by the plane is 480.6m.

(b)

To determine

The vertical velocity should the supplies be given so that they arrive precisely.

Answer to Problem 35P

The magnitude of the vertical velocity of the supplies is 8.41m/s.

Explanation of Solution

Given Data:

The horizontal distance is 425.

Horizontal distance, R = 425

Calculation:

The kinematic equation for the plane:

  $\begin{array}{l}y-y_o=v_y^{\text{'}}t\text{'}+\frac{1}{2}at{\text{'}}^2\\ \text{where,}\\ y_o=\text{initial height}\\ y=\text{final height}\\ a=\text{acceleration}\\ t\text{'}=\text{time}\\ v{\text{'}}_y=\text{vertical velocity}\end{array}$

As the supply is moving against the gravitational field, then the acceleration will be opposite to the gravitation.

  $a=-g$

Then, the equation is expressed as:

  $\begin{array}{l}y-y_o=v{\text{'}}_{y}t\text{'}+\frac{1}{2}at{\text{'}}^2\\ y-y_o=v{\text{'}}_{y}t\text{'}+\frac{1}{2}\left(-g\right)t{\text{'}}^2\\ y-y_o=v{\text{'}}_{y}t\text{'}-\frac{1}{2}gt{\text{'}}^2\end{array}$

The horizontal velocity:

  $v_x=\frac{x\text{'}}{t\text{'}}$

Rewriting the above equation for t:

  $t\text{'}=\frac{x\text{'}}{v_x}$

Substituting the known value into the above equation:

  $\begin{array}{l}t\text{'}=\frac{425}{69.4\text{m/s}}\\ t\text{'}=6.12\text{s}\end{array}$

Rewriting the equation:

  $v{\text{'}}_y=\left(y-y_o+\frac{1}{2}gt{\text{'}}^2\right)\left(\frac{1}{t\text{'}}\right)$

Substituting the known value into the above equation:

  $\begin{array}{l}v{\text{'}}_y=\left(0-235+\left(\frac{1}{2}\right)\left(9.8{\text{m/s}}^2\right){\left(6.12\text{s}\right)}^2\right)\left(\frac{1}{6.12}\right)\\ v{\text{'}}_y=-8.41\text{m/s}\end{array}$

Here, the negative sign indicates the velocity is moving downward and it is against the gravity.

The magnitude of the vertical:

  $\left|v{\text{'}}_y\right|=8.41\text{m/s}$

Therefore, the magnitude of the vertical velocity of the supplies is 8.41m/s.

(c)

To determine

The speed of the supplies land in the latter case.

Answer to Problem 35P

The speed of the supplies land in the latter case is 97.4m/s.

Explanation of Solution

Given Data:

The height of plane is 235 m.

Horizontal distance R = 425

Calculation:

The horizontal velocity:

  $v_x=69.4\text{m/s}$

The vertical velocity:

  $v_y=v{\text{'}}_y-gt\text{'}$

Substituting the known values into the above equation:

  $\begin{array}{l}{v}_y=-8.41\text{m/s-}\left(\text{9}{\text{.8m/s}}^{\text{2}}\right)\left(6.12\text{s}\right)\\ v_y=-68.34\text{m/s}\end{array}$

The magnitude of the net velocity:

  $\begin{array}{l}{v}_y=-8.41\text{m/s-}\left(9.{\text{8m/s}}^{\text{2}}\right)\left(6.12\text{s}\right)\\ =-68.34\text{m/s}\end{array}$

The magnitude of the net velocity of the plane:

  $\begin{array}{l}{v}_{net}=\sqrt{v_x^2+v_y^2}\\ =\sqrt{{\left(69.8\text{m/s}\right)}^2+{\left(-68.34\text{m/s}\right)}^2}\\ =97.4\text{m/s}\end{array}$

Hence, the speed of the supplies land in the latter case is 97.4m/s.