Question

A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 130 m above the glacier at a speed of 110 m/s . You may want to review (Page) . For help with math skills, you may want to review: Mathematical Expressions Involving Squares For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Dock jumping. How far short of the target should it drop the package?

Answer 1

Answer:

$t =\sqrt{\frac{2*(-100m)}{-9.8 m/s^2}}= 4.518 s$

And now we can find the final distance on the x axis using the formula:

$D = V_x t$

The velocity on x not changes and is the same plane spped $V_x = 150 m/s$, if we replace we got:

$X= 150 m/s * 4.518 s= 677.63 m$

Explanation:

For this case we have a illustration for the problem in the figure attached.

We need to find how far short of the target should be the plane, we have the following info given:

$v_{ix}= 150 m/s$

$v_{iy}=0 m/s$

$y_{f} =-100m$

$y_i =0m$

$x_i = 0m$

How far short of the target should it drop the package?

First we can find the total time in order to reach the groung using the following kinematic formula:

$y_f = y_i + v_{iy} +\frac{1}{2}gt^2$

And replacing we have:

$-100 m = 0 +0 -\frac{1}{2} (9.8 m/s^2) t^2$

And solving for t we got:

$t =\sqrt{\frac{2*(-100m)}{-9.8 m/s^2}}= 4.518 s$

And now we can find the final distance on the x axis using the formula:

$D = V_x t$

The velocity on x not changes and is the same plane spped $V_x = 150 m/s$, if we replace we got:

$X= 150 m/s * 4.518 s= 677.63 m$