Question

The St. Louis Arch has a height of 192 m. Suppose that a stunt woman of mass 84 kg jumps off the top of the arch with an elastic band attached to her feet. She reaches the ground at zero speed. The acceleration of gravity is 9.81 m/s 2 . Find her kinetic energy after 2.6 s of the flight. Assume the elastic band has no length and obeys Hooke’s Law. Answer in units of kJ

Answer 1

To solve this problem it is necessary to apply the kinematic equations of motion for speed and distance, as well as the concepts related to kinetic energy.

The change in the height of a body subject to gravity is given by

$h = \frac{1}{2} gt^2 \rightarrow t = \sqrt{\frac{2h}{g}}$

Where

h = Height

g =Gravity

t = time

Replacing with our values we have that the time is

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(192)}{9.8}}$

$t = 6.25s$

From speed as a function of change between acceleration and time we have then that after 2.6 seconds the speed would be

$g = \frac{v}{t} \rightarrow v = g*t$

$v = 9.8*2.6$

$v = 25.48m/s$

The kinetic energy would be given by

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (84)(25.48)$

$KE = 1070.16J$

Therefore the kinetic energy after 2.6s is 1070.16J