Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 m/s), (a) how far in advance of the recipients (horizontal distance) must the goods be dropped. (b) Suppose, instead that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position. (c) with what speed do the supplies land in the latter case?

Answer 1

Concepts and reason

The concepts used in this problem are Newton’s equations of motion and impact speed. Solve the first and second part of the problem using Newton’s equations of motion and then find the impact velocity by calculating the vertical speed.

Fundamentals

Newton’s equations of motion:

The Newton’s equations of motion are used to find the initial speed, final speed, acceleration, distance and time of a motion. These equations describe the motion of an object as a function of time. These are the equations of kinematics.

Three equations of motion are:

$\begin{array}{l}\\v - u = at\\\\s = ut + \frac{1}{2}a{t^2}\\\\{v^2} - {u^2} = 2as\\\end{array}$

Here,

$\begin{array}{l}\\v{\rm{ is the final speed}}{\rm{.}}\\\\u{\rm{ is the initial speed}}{\rm{.}}\\\\a{\rm{ is acceleration}}{\rm{.}}\\\\s{\rm{ is distance}}{\rm{.}}\\\\t{\rm{ is time}}{\rm{.}}\\\end{array}$

Impact Velocity:

The impact speed is the maximum speed of an object just before it comes to rest. If the motion is projectile motion, then there are two components of velocity before the object comes to rest. Impact velocity is the resultant of x-component and y-component of velocity.

$v' = \sqrt {{v_x}^2 + {v_y}^2}$

Here,

$\begin{array}{l}\\{v_x}{\rm{ is the x - component of velocity}}{\rm{.}}\\\\{v_y}{\rm{ is the y - component of velocity}}{\rm{.}}\\\end{array}$

(a)

First, calculate the time using second equation of motion.

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

Here, the given data is:

$\begin{array}{l}\\{\rm{The initial speed is }}u = {\rm{0}}{\rm{.}}\\\\{\rm{The distance is }}s = {\rm{235 m}}{\rm{.}}\\\\{\rm{The acceleration is }}a = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\\end{array}$

$\begin{array}{c}\\s = ut + \frac{1}{2}a{t^2}\\\\235\;{\rm{m}} = \left( 0 \right)t + \frac{1}{2}\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right){t^2}\\\\{t^2} = \frac{{235\;{\rm{m}}}}{{4.9\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}\\\\t = \sqrt {47.95\;{{\rm{s}}^{\rm{2}}}} \\\\ = 6.92\;{\rm{s}}\\\end{array}$

The final speed when the goods drop is 69.4 m/s.

The distance covered is:

$\begin{array}{c}\\s' = vt\\\\ = \left( {69.4\;{\rm{m/s}}} \right)\left( {6.92\;{\rm{s}}} \right)\\\\ = {\bf{480}}{\bf{.25}}\;{\bf{m}}\\\end{array}$

(b)

The time taken by supplies is:

$t = \frac{{{s_x}}}{{{v_x}}}$

Here, ${s_x}{\rm{ is the horizontal distance and }}{v_x}{\rm{ is the horizontal speed}}{\rm{.}}$

$\begin{array}{c}\\t = \frac{{425\;{\rm{m}}}}{{69.4\;{\rm{m/s}}}}\\\\ = 6.12\;{\rm{s}}\\\end{array}$

Use the second equation of motion to find the vertical speed:

$\begin{array}{c}\\{s_y} = {v_y}t + \frac{1}{2}g{t^2}\\\\ - 235\;{\rm{m}} = {v_y}\left( {6.12\;{\rm{s}}} \right) + \frac{1}{2}\left( { - 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right){\left( {6.12\;{\rm{s}}} \right)^2}\\\\{v_y} = \frac{{ - 235\;{\rm{m}} + 183.5\;{\rm{m}}}}{{6.12\;{\rm{s}}}}\\\\ = - {\bf{8}}{\bf{.42}}\;{\bf{m/s}}\\\end{array}$

(c)

The above calculated vertical speed is the initial speed for this motion.

Use first equation of motion:

$v - u = at$

$\begin{array}{c}\\{\rm{Substitute the values:}}\\\\{v_y}' - {v_y} = - gt\\\\{v_y}' = {v_y} - gt\\\\ = \left( { - 8.42\;{\rm{m/s}}} \right) - \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {6.12\;{\rm{s}}} \right)\\\\ = - 68.4\;{\rm{m/s}}\\\end{array}$

The final speed or the resultant speed is:

$\begin{array}{c}\\v = \sqrt {{v_x}^2 + {v_y}{'^2}} \\\\ = \sqrt {{{\left( {69.4\;{\rm{m/s}}} \right)}^2} + {{\left( {68.4\;{\rm{m/s}}} \right)}^2}} \\\\ = {\bf{97}}{\bf{.4}}\;{\bf{m/s}}\\\end{array}$

Ans: Part a

The goods dropped at a distance of 480.25 m.

Part b

The vertical speed is 8.42 m/s and downwards.

Part c

The speed of landing of supplies is 97.4 m/s.