Question

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 18 m/s and is flying 225 m above the target, how far away (horizontally) from the target must the rescue pilot drop the package?

Answer 1

Answer:

x = 121.86 m

Explanation:

It is given that,

The vertical velocity of a plane, $v_x=18\ m/s$

It is flying above the target, h = 225 m

First we will find the time taken by the plane to reach the ground. It can be calculated using the second equation of motion as follow :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 225}{9.81}} \\\\t=6.77\ s$

Let x is the horizontal distance from the target. It can be calculated as follows :

$x=v_xt\\\\x=18\times 6.77\\\\x=121.86\ m$

So, the rescue pilot drop the package at a distance of 121.86 m from the target.