Question

A student is determined to test the law gravity for himself by walking off a skyscraper 900 ft high, with a stopwatch on hand, and starts his free fall (zero initial velocity). Five seconds later, superman arrives at the scene and dives off the roof to save the student. What must superman’s initial velocity be in order that he catches the student just before the ground is reached? A. 315 ft/sec C. 320 ft/sec B. 300 ft/sec D. 350 ft/sec

Answer 1

Explanation:

Correct option is A)

Height of the skyscapper, s=320 m

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

s=ut+21at2

As the initial velocity u=0and a=10m/s2

 then, 

320=0+21×10×t2

320=21×10t2

t2=10640=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/s2 and t=3 we need to find the initial velocity u.

320=(u×3)+21×10×3×3

320=3u+45

320−45=3u

275=3u

u=3275⇒91.66m

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