Question

A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 88m above the glacier at a speed of 150m/s. How far short of the target should it drop the package?

________m

Answer 1

Concepts and reason

The main concept used is kinematics equations.

Initially, calculate the time taken by using the kinematics equation in vertical direction. Finally calculate the distance the package dropped by using the kinematics equation in horizontal direction.

Fundamentals

From the linear kinematic expression, the expression of distance covered by the object is as follows,

$s = vt$

Here,$s$is the distance covered by the object,$v$is the velocity and $t$is the time.

From the linear kinematic expression the expression of distance covered by the object in terms of the acceleration due to the gravity is expressed as follows,

$s = ut + \frac{1}{2}g{t^2}$

Here,$u$is the initial velocity and $g$is the acceleration due to the gravity.

Calculate the time taken to reach the package in Greenland.

The expression of the vertical distance in terms of the acceleration due to the gravity is equal to,

$s = ut + \frac{1}{2}g{t^2}$

Substitute $0$for$u$,$88{\rm{ m}}$for$s$and $9.8{\rm{ m/}}{{\rm{s}}^2}$for$g$in the above expression of the vertical distance,

$88{\rm{ m}} = 0 + \left( {\frac{1}{2}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {{t^2}} \right)$

Rearrange the above expression in terms of the time$t$.

${t^2} = \frac{{\left( 2 \right)\left( {88{\rm{ m}}} \right)}}{{\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}$

Taken square root both sides,

$\begin{array}{c}\\t = \sqrt {\frac{{\left( 2 \right)\left( {88{\rm{ m}}} \right)}}{{\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}} \\\\ = 4.24{\rm{ s}}\\\end{array}$

Calculate the distance at which the package is drop at the target.

The expression of the distance at which the package is drop at the target.

$s = vt$

Substitute $150{\rm{ m/s}}$for$v$and $4.24{\rm{ s}}$for$t$in the above expression of the $s$

$\begin{array}{c}\\s = \left( {150{\rm{ m/s}}} \right)\left( {4.24{\rm{ s}}} \right)\\\\ = 636{\rm{ m}}\\\end{array}$

Ans:

The distance at which the package is drop at the target is equal to$636{\rm{ m}}$.