Superman...

Classical Mechanics Level 2

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. 5 sec 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 any freely falling body.] (g=10 m/s²)

187/2 m/s 275/3 m/s it is not possible 98 m/s

Manan Hingorani by Manan Hingorani , 23, India

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1 solution

Ankita Nayak
May 16, 2015

Displacement (s) = 320 m / s 2 320 m/s^{2}

S = u t + 1 2 a t 2 S = ut + \frac{1}{2}at^{2}

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

t = 8 s t= 8 s

Superman arrives 5 s later. Thus, he has only 8 − 5 = 3 s 8-5=3 s to catch the student. Since Superman catches the student just before reaching the ground, he also falls through 320 m. 320 = 3 × u + 1 2 × 10 × 3 2 320= 3\times u +\frac{1}{2}\times 10\times 3^{2}

u = 275 3 m / s \boxed {u= \frac{275}{3} m/s}

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