Question

PLS HELP MY TOPIC IS ABOUT PROJECTILE MOTION

2. A rescue pilot wishes to drop a package of emergency supplies so that it lands as close as possible to a target. If the plane travels with a velocity of 81 m/s and is flying 125 m above
the target, how far away (horizontally) from the target must the rescue pilot drop the package?

Answer 1

Do you have the answer to check? Cause I got 12.36m away from the target

Answer 2

package velocity = airplane velocity , when its droped

Answer 3

from the airplane

Answer 4

AJ can you work this problem out? I want to check if my answer is right/not

Answer 5

You drop a package from an airplane and once it hits ground the velocity of the object is zero. Let's analyze this problem. Before the package is dropped the horizontal velocity of the package is the same as the plane because its on the plane. The vertical velocity is zero because the package is not going down. Since where the package will land depend how long the package stays in the air it is reasonable we need to find time. How long it stays in the air before it hits the ground.

Lets put our words into math variables. We know the following:
Vxi = 81 m/s
Vyi = 0 m/s
yi= 125 m
yf= 0 m
g = 9.8 m/s^2

The equations will use \[y _{f}= y _{i} + v _{i}t - \frac{ 1 }{ 2 }g t^{2}\] and\[x=v _{i}t\] Y is for the vertical and x for the horizontal. First we need to find time, how long the package stays in the air. That depends on the vertical direction which is the y coordinate. Now plug in the values we know for y.

\[0 = 125 + 0 -\frac{ 1 }{ 2 }(9.8)t ^{2}\]After doing a little algebra we get t = 5.05 s. So that means the package stays in the air for 5.05 seconds. Now we can find how far way from the target horizontally.\[x = (81)(5.05)\] So x =409.05 m. So the rescue pilot must drop the package 409.05 m away so it can land at the correct location.