Question

A student determined to test a law of gravity for him walks off a skyscraper \[320m\] high with a stopwatch in hand and starts his free fall (zero initial velocity). 5 seconds later, superman arrives at the scene and dives off the roof to save the student. What must be superman’s initial velocity in order that he catches the student just before reaching the ground? [Assume that the superman’s acceleration is that of a free falling body, \[g=10m/{{s}^{2}}\]].
A. \[91.66m{{s}^{-1}}\]
B. \[25.8m{{s}^{-1}}\]
C. \[4.785m{{s}^{-1}}\]
D. Cannot be determined

Answer 1

Hint: When a body is dropped from some height then initial velocity of the body is zero and the acceleration of the body is equal to the acceleration due to gravity. The motion of the body during free fall is uniform accelerated motion. As the motion of the body is uniformly accelerated, hence the equation of motion can be applied to calculate the unknown terms.

Complete step by step solution:
Height of the skyscraper, $H=320m$
According to equations of the motion, the student having a free fall can be described by the following equation.
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Where,
$s=$Displacement
$u=$Initial velocity$=0m/s$
$a=$Acceleration$=10m/{{s}^{2}}$
$t=$Time taken by the body
Then,
Putting the values in the equation of motion,
$\begin{align}
  & 320=\left( 0 \right)t+\dfrac{1}{2}\left( 10 \right){{t}^{2}} \\
 & {{t}^{2}}=64 \\
 & t=8s \\
\end{align}$
So, the total time taken by the student for the free fall from the skyscraper is 8s.
It is given that Superman arrives after 5 s, so the time taken by Superman to reach the bottom of the skyscraper to save the student is
$\left( 8-5 \right)s=3s$
Superman has taken 3s to cover the total height of the skyscraper.
Again, using the equation of motion $s=ut+\dfrac{1}{2}a{{t}^{2}}$
Assuming $u$as the initial velocity of Superman
$\begin{align}
  & 320=3u+\dfrac{1}{2}\left( 10 \right){{\left( 3 \right)}^{2}} \\
 & 3u=320-45 \\
 & u=\dfrac{275}{3}m/s \\
 & =91.67m/s
\end{align}$

Therefore, to save the student the superman must have velocity equal or greater than 91.67 m/s.

Note: - It is assumed that the acceleration due to gravity is uniform on the surface of the earth.
- It is assumed that the air resistance is negligible.