Question

1. a ski jumper competing for an olympic gold medal wants to jump a horizontal distance of 149 meters. the takeoff point of the ski jump is at a height of 38.0 meters. with what horizontal velocity must he leave the jump in order to travel 149 meters?

a. 50.2 m/s
b. 59.2 m/s
c. 53.5 m/s
d. 55.3 m/s

Explanation:

Step1: Calculate the time of flight

The vertical motion is a free - fall motion. Using the equation \(h = v_{0y}t+\frac{1}{2}gt^{2}\), where \(v_{0y} = 0\) (initial vertical velocity), \(h = 38.0\ m\) and \(g=9.8\ m/s^{2}\).
So, \(h=\frac{1}{2}gt^{2}\), then \(t=\sqrt{\frac{2h}{g}}\)
Substitute \(h = 38.0\ m\) and \(g = 9.8\ m/s^{2}\) into the formula:
\(t=\sqrt{\frac{2\times38.0}{9.8}}\approx\sqrt{\frac{76}{9.8}}\approx 2.79\ s\)

Step2: Calculate the horizontal velocity

In horizontal motion (no - acceleration, \(a_x = 0\)), the horizontal distance \(x = v_{0x}t\), then \(v_{0x}=\frac{x}{t}\)
Given \(x = 149\ m\) and \(t\approx2.79\ s\)
\(v_{0x}=\frac{149}{2.79}\approx53.5\ m/s\)

Answer:

c. \(53.5\ m/s\)