Question

the record for a ski jump is 180 m set in 1989. Assume the jumper comes off the end of the ski jump horizontally and falls 90 m vertically before contacting the ground . what was the initial horizontal speed of the jumper?

Answer 1

Let's write the equations of motion on both x- (horizontal) and y- (vertical) axis. On the x-axis, it's a uniform motion with constant velocity vx. On the y-axis, it is a uniformly accelerated motion with initial height h=90 m and acceleration of $g=9.81 m/s^2$ pointing down (so with a negative sign):
$S_x(t)=v_x t$
$S_y(t)=h- \frac{1}{2} gt^2$

First, let's find the time at which the jumper reaches the ground. This happens when Sy(t)=0:
$0=h- \frac{1}{2} g t^2$
and so 
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 90m}{9.81 m/s^2} }=4.28 s$

Then, we can find the horizontal speed. In fact, we know that at the time t=4.28 s, when the jumper reached the ground, he covered exactly 180 m, so Sx=180 m. Using this into the law of motion in x, we find
$v_x= \frac{S_x}{t} = \frac{180 m}{4.28 s} =42 m/s$