Question

Suppose the rescue plane of Problem 31 releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers’ position (Fig. 3–39)? With what speed do the supplies land?

FIGURE 3-39Problem 32

Short answer

The supplies should be given a vertical velocity of$8.41\mathrm{m}/\mathrm{s}$in the downward direction so that they arrive precisely at the climber’s position. The supplies land at the climber’s position at the speed of $97.44\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Step 1. Kinematic equations for projectile motion

When a projectile is projected horizontally from a height yabove the ground with an initial velocity$v_{x0}$, it falls downward in a parabolic path under the effect of gravity. If the initial point is taken as the origin and downward direction is taken as positive y-axis, then the kinematic equations for projectile motion will be as follows:

• The kinematic equations for thehorizontal motion of projectile are:

$\begin{array}{c}{v}_x=v_{x0}\\ x=v_{x0}t\end{array}$

• The kinematic equations for thevertical motion of the projectile are:

$\begin{array}{c}{v}_y=v_{y0}+\mathrm{g}t\\ y=v_{y0}t+\frac{1}{2}\mathrm{g}{t}^2\\ v_y^2=v_{y0}^2+2\mathrm{g}y\end{array}$

Here, xandy are the horizontal and vertical displacements of the projectile traveled in time t.

Step 2. Given information: 

Consider that the supplies which are thrown by the plane are projectiles.

The horizontal distance of the projectile is $x=425\mathrm{m}$.

The vertical distance of the projectile is $y=235\mathrm{m}$.

The initial horizontal velocity of the projectile will be the same as that of the plane, i.e., $v_{x0}=69.4\mathrm{m}/\mathrm{s}$.

Let the initial vertical velocity of the projectile be $v_{y0}$and time taken by the projectile to reach the ground be t.

Step 3. Determination of time taken by the supplies to reach the ground

Using the kinematic equation for the horizontal motion of the projectile, you will get time t as:

$\begin{array}{c}x=v_{x0}t\\ t=\frac{x}{v_{x0}}\\ =\frac{\left(425\mathrm{m}\right)}{\left(69.4\mathrm{m}/\mathrm{s}\right)}\\ =6.12\mathrm{s}\end{array}$

Step 4. Determination of the initial vertical velocity of the supplies

Using the kinematic equation for vertical motion of the projectile, you will get:

$\begin{array}{c}y=v_{y0}t+\frac{1}{2}\mathrm{g}{t}^2\\ 235\mathrm{m}=v_{y0}\left(6.12\mathrm{s}\right)+\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(6.12\mathrm{s}\right)}^2\\ v_{y0}=\frac{\left(235\mathrm{m}\right)}{\left(6.12\mathrm{s}\right)}-\frac{1}{2}\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(6.12\mathrm{s}\right)}^2}{\left(6.12\mathrm{s}\right)}\\ =38.40\mathrm{m}/\mathrm{s}-29.99\mathrm{m}/\mathrm{s}\\ =8.41\mathrm{m}/\mathrm{s}\end{array}$

Thus, the supplies should be dropped by giving an initial vertical velocity of$8.41\mathrm{m}/\mathrm{s}$ in a downward direction.

Step 5. Determination of the speed of the supplies with which they land at the climber’s position

Using the kinematic equation for the vertical motion of projectile, you will get:

$\begin{array}{c}{v}_y=v_{y0}+\mathrm{g}t\\ =\left(8.41\mathrm{m}/\mathrm{s}\right)+\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(6.12\mathrm{s}\right)\\ =68.39\mathrm{m}/\mathrm{s}\end{array}$

The magnitude of the final velocity of the supplies, i.e., final speed of the supplies, will be:

$\begin{array}{c}v=\sqrt{v_x^2+v_y^2}\\ =\sqrt{{\left(69.4\mathrm{m}/\mathrm{s}\right)}^2+{\left(68.39\mathrm{m}/\mathrm{s}\right)}^2}\\ =\sqrt{9493.55}\\ =97.44\mathrm{m}/\mathrm{s}\end{array}$

Thus, the supplies land on the ground with a speed of $97.44\mathrm{m}/\mathrm{s}$.