A student determined to test the law of gravity for himself walks off a skyscraper 900 ft high, stopwatch in 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. (a) What must Superman's initial velocity be in order that he catch the student just before the ground is reached? (b) What must be the height of the skyscraper so that even Superman can't save him? (Assume that Superman's acceleration is that of any freely falling body.)

Question

Alani

Physics

29 July, 22:05

A student determined to test the law of gravity for himself walks off a skyscraper 900 ft high, stopwatch in 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. (a) What must Superman's initial velocity be in order that he catch the student just before the ground is reached? (b) What must be the height of the skyscraper so that even Superman can't save him? (Assume that Superman's acceleration is that of any freely falling body.)

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Answers (1)

Know the Answer?

Payton Rivers · Answer 1

a) 323.517 ft/s

b) 402.5 ft

Explanation:

At t = 5s the student had travelled:

y = (32/2) * 5^2 ft

y = 402.5 ft

The time at which the student will reach the ground is given by:

900 ft = (32/2 ft/s^2) * t^2

Solving for t

t = (1800/32 s^2) ^ (1/2)

t = 7.47667 s

Therefore Superman need to travell 900 ft in 2.47667 s

The equation of motion of Superman will be:

ys (t) = v0*t + (32.2/2 ft/s^2) * t^2

We are looking for a velocity v0 such that ys (2.47667 s) = 900 ft:

900 ft = v0 * (2.47667 s) + (32.2/2 * (2.47667) ^2) ft

v0 = (900 - (32.2/2) (2.47667) ^2) / (2.47667) ft/s

v0 = 323.517 ft/s

The hight of the skyscraper so that even Superman cant save him will be the same as the distance y0 the student has travelled when superman arrived, that is:

y (t) = - (32.2 (ft/s^2) / 2) * (25s^2) = 402.5 ft