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

x = 121.86 m

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

Here, u = 0 and a = g

`h=(1)/(2)gt^2\\\\t=\sqrt{(2h)/(g)} \\\\t=\sqrt{(2* 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* 6.77\\\\x=121.86\ m`

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