Question

1. a person who has a mass of 65 kg goes on the sky tower ride at an amusement park. the ride is simply a free fall from the top of a tower into a net below. if the person reaches a final velocity of 24.6 m/s just before hitting the net, from what height did the rider drop? ignore friction with the air.

Explanation:

Step1: Select kinematic equation

We use the kinematic equation for free fall (initial velocity $v_0=0$, acceleration $a=g=9.8\ \text{m/s}^2$):
$$v^2 = v_0^2 + 2gh$$

Step2: Isolate height $h$

Rearrange the equation to solve for $h$:
$$h = \frac{v^2 - v_0^2}{2g}$$

Step3: Substitute given values

Substitute $v=24.6\ \text{m/s}$, $v_0=0$, $g=9.8\ \text{m/s}^2$:
$$h = \frac{(24.6)^2 - 0^2}{2\times9.8} = \frac{605.16}{19.6}$$

Answer:

$30.9\ \text{m}$ (rounded to one decimal place)