Question

The Jurassic Park ride at Universal Studios theme park drops 25.6 m straight down essentially from rest. Find the time for the drop and the velocity at the bottom.

Answer 1

Answer:

V=22.4m/s;T=2.29s

Explanation:

We will use two formulas in order to solve this problem. To determine the velocity at the bottom we can use potential and kinetic energy to solve for the velocity and use the uniformly accelerated displacement formula:

$mgh=\frac{1}{2}mv^{2}\\\\X= V_{0}t-\frac{gt^{2}}{2}$

Solving for velocity using equation 1:

$mgh=\frac{1}{2}mv^{2} \\\\gh=\frac{v^{2}}{2}\\\\\sqrt{2gh}=v\\\\v=\sqrt{2*9.8\frac{m}{s^2}*25.6m}=22.4\frac{m}{s}$

Solving for time in equation 2:

$-25.6m = 0\frac{m}{s}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\-51.2m=-9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{\frac{51.2m}{9.8\frac{m}{s^{2}}}}=2.29s$