Question

A person standing on the top of a cliff 171ft high has to throw a packet to his friend standing on the ground $228ft$ horizontally away. If he throws the packet directly aiming at the friend with a speed of $15.0ft/s$ , how short will the packet fall?

Answer 1

Hint: The question is based on two-dimensional kinematics. Knowing how to break velocity and acceleration into horizontal and vertical components will help in answering this problem. Additionally, if you know the formulas of various quantities such as height, velocity, acceleration, etc. for two-dimensional kinematics, it is helpful.

Complete answer:

Let’s start by constructing a diagram of the problem. We have a cliff of height $171ft$ , from where a packet is thrown horizontally to a point on the ground $228ft$ away with velocity $15.0ft/s$ . The gravity $\left( {g = 32ft/{s^2}} \right)$ acts vertically downwards. This leads to the ball making a trajectory instead of following the straight path. We will consider the positive x-axis as it moves from left to right and for the y-axis, positive is from the top towards the bottom. Hence, $g$ is positive. Diagrammatically, we have the following scenario.

The straight route distance from $P$ to $Q$ is $\sqrt {{{171}^2} + {{228}^2}} = 285ft$ .

The packet is thrown along the direction $PQ$ with velocity $15ft/s$ . If we resolve this velocity into vertical and horizontal components;

The vertical component of the velocity ${v_y}$ is,

${v_y} = 15\sin \theta $

$ \Rightarrow {v_y} = 15 \times \dfrac{{PO}}{{PQ}}$

On putting the values we get,

${v_y} = 15 \times \dfrac{{171}}{{285}} = 9ft/s$ ............ $\left( 1 \right)$

The horizontal component of the velocity ${v_x}$ is,

${v_x} = 15\cos \theta $

$ \Rightarrow {v_x} = 15 \times \dfrac{{OQ}}{{PQ}}$

On putting the values we get,

${v_x} = 15 \times \dfrac{{228}}{{285}} = 12ft/s$

The time to touch the ground for the packet is decided by gravity.

Hence we will determine the time taken to touch the ground using vertical component

We need to use the second equation of kinematics i.e.

$S = ut + \left( {\dfrac{1}{2}} \right)g{t^2}$

$ \Rightarrow 171 = 9 \times t + \left( {\dfrac{1}{2}} \right) \times 32 \times {t^2}$

On simplifying the above equation we get,

$171 = 9t + 16{t^2}$

$ \Rightarrow 16{t^2} + 9t - 171 = 0$

On solving this quadratic equation for $t$ we get the values

$t = 3,\left( { - \dfrac{{57}}{{16}}} \right)$

As time cannot be negative, the time to touch the ground is $t = 3s$ .

Hence the horizontal distance covered by the packet will be,

$OQ' = {v_x} \times t$

Substituting the values in the above equation we get,

$OQ' = 12 \times 3 = 36ft$

So, the packet will fall short of $Q$ is,

$QQ' = OQ - OQ'$

$ \Rightarrow QQ' = 228 - 36 = 192ft$

Therefore, the packet will fall at a distance of $36ft$ from $O$, short of $192ft$ to $Q$.

Note:
The velocity of the projectile keeps varying along its trajectory. As the projectile rises above the ground, its velocity keeps on reducing. At the highest point, velocity is minimum . As the projectile starts falling, its velocity starts increasing again.