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- (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- (1)/(2) g t^2`
and so

`t= \sqrt{ (2h)/(g) }= \sqrt{ (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= (S_x)/(t) = (180 m)/(4.28 s) =42 m/s`