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

`t =\sqrt{(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`

Step-by-step 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) +(1)/(2)gt^2`

And replacing we have:

`-100 m = 0 +0 -(1)/(2) (9.8 m/s^2) t^2`

And solving for t we got:

`t =\sqrt{(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`