Question

A student determined to test the law of gravity for himself walks off a sky scraper 320 m high with a stopwatch in hand and starts his free fall (zero initial velocity). 5 second 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 can save the student. [Assume that the superman's acceleration is that of any freely falling body.] $(g=10m/s^2)$

• A. $98$ m/s
• B. $\frac{275}{3}$ m/s
• C. $\frac{187}{2}$ m/s
• D. It is not possible.

Answer 1

The correct option is B $\frac{275}{3}$ m/s

Time taken by the student to reach ground:

$320=\frac{1}{2}g{t}^2$
$\Rightarrow t=8sec$

Thus, the superman has to move for $(8-5)$$=3sec$ to save the student.
Now, $h=ut-\frac{1}{2}g{t}^2$
$320=3u+\frac{1}{2}\times 10\times 9$
$u=91.66=\frac{275}{3}m{s}^{-1}$
Hence, option B is correct.