Question

A healthy young man standing at a distance of $7m$ from a $11.8m$ high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height $1.8m$?

Answer 1

Hint: You can begin by first calculating the amount of time the kid will take to reach the man at the bottom of the building. Keep in mind that the kid finally reaches the man’s arm height, so account for that in your estimation of the vertical distance that the kid travels. The time that you calculate for the kid is also the time that the man must take to cover the horizontal distance to the bottom of the building. Finding the velocity will now be pretty straightforward.

Formula used: Kinematic equation of motion:
$S=ut+{\displaystyle \frac{1}{2}}g{t}^2$, where, S is the distance travelled, u is the initial velocity, t is the time taken, and g is the acceleration due to gravity.
Velocity $v={\displaystyle \frac{distance}{time}}$

Complete step by step answer:
We are required to find the speed with which the man must run so that he reaches the bottom of the building just in time to save the kid.
We already know that he has to travel a distance of $7m$, so all that we need to do find out is the time he takes to cover that distance. In order to save the kid, he needs to cover the $7m$ distance in at least the same amount of time the kid takes to reach his arm’s height at the bottom of the building.

We can thus find the time the kid takes to slip all the way to the bottom of the building from a kinematic equation of motion:
$S=ut+{\displaystyle \frac{1}{2}}g{t}^2$, where,
$u$ is the initial velocity with which the kid falls $u=0m{s}^{-1}$,
$S$ is the distance through which the kid falls until he reaches the man’s arm height $S=11.8–1.8=10m$,
$g$ is the acceleration with which the kid falls due to gravity $g=9.8m{s}^{-2}$,
$t$ is the time the kid takes to travel distance $S$ which we can now calculate:
$10=0\times t+{\displaystyle \frac{1}{2}}\times 9.8\times t^2\Rightarrow t^2={\displaystyle \frac{10}{4.9}}=2.04s\Rightarrow t=\sqrt{2.04}=1.43s$
Now, this will also be the maximum amount of time that the man can take to reach the bottom of the building. Therefore, for the man has to run with a velocity of:
$v={\displaystyle \frac{d}{t}}={\displaystyle \frac{7}{1.43}}=4.9m{s}^{-1}$

Therefore, in order to catch the kid, the man has to run with a velocity of at least $4.9m{s}^{-1}$

Note: In the problem, do not overlook the fact that the kid is caught at the man’s arms height since it is only the man that will cover a horizontal distance to reach the bottom of the building while the kid is not necessarily caught at the bottom of the building. Note that the velocity that we obtain is the minimum velocity with which the man should run to be able to save the kid, and the time he takes is the maximum window that is available to save the kid, beyond which taking more time will not get him to the bottom of the building before the kid hits the ground.