Question

A student determined to test the law of gravity for himself 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 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

The correct option is A $91.66m{s}^{-1}$

Height of the skyscapper, s=$320m$

so according to equations of the motion, the student having a free fall can be described by the following equation.

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

As the initial velocity $u=0$and $a=10m/s^2$

then,

$320=0+\frac{1}{2}\times 10\times t^2$

$320=\frac{1}{2}\times 10t^2$

$t^2=\frac{640}{10}=64$

$t=8s$

so the total time taken for the free fall of the student from the skyscapper is 8s.

According to the question, superman arrives after $5$ s, so he has $(8-5)=3s$ to save the student from reaching the ground.

Again using the same equation of the motion, by substituting $s=320m$, $a=10m/s^2$ and $t=3$ we need to find the initial velocity $u$.

$320=(u\times 3)+\frac{1}{2}\times 10\times 3\times 3$

$320=3u+45$

$320-45=3u$

$275=3u$

$u=\frac{275}{3}\Rightarrow 91.66m$

the initial velocity of the superman should be $91.66m/s$