Question

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m 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.Superman leaves the roof with an initial speed $v_0$ that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of $v_0$ be so that the Superman catches the student just before they reach the ground?
b.On the same graph, sketch the positions of the student and of the Superman as functions of time. Take Supermans initial speed to have the value calculated in part (a).
c. If the height of the skyscraper is less than some minimum value, even the Superman cannot reach the student before he hits the ground, what is this minimum height?

Answer 1

a. Suppose that Superman falls for a time t and that the student has been falling for a time f0 before Supermans leap (in this case, $t_0$=5 s). Then the height h of the building is related to t and $t_0$ in two different ways:

- h = $v_{0y}t-\left(1/2\right)g{t}^2$ = - (1/2)g (t + $t_0{)}^2$ ...(i)

where $v_{0y}$ is Supermans initial velocity. Solving the second

t gives t = $\sqrt{\frac{2h}{g}}$ - $t_0=\sqrt{\frac{2\times 180}{10}}$ - 5 = 1 s

Solving the first for $v_{0y}$ gives $v_{0}y=-\frac{h}{t}+\frac{g}{2}t$ and the substitution of numerical values gives t = 1.0 s and $v_{0y}$ = -175 m/s, with minus sign indicating a downward initial velocity.

(b). Refer figure.

(c).If the skyscraper is so short that the student is already on the ground, then h = (1/2)g$t_0^2$ = 125 m.