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 [tex]g=9.81 m/s^2[/tex] pointing down (so with a negative sign):
[tex]S_x(t)=v_x t[/tex]
[tex]S_y(t)=h- \frac{1}{2} gt^2[/tex]
First, let's find the time at which the jumper reaches the ground. This happens when Sy(t)=0:
[tex]0=h- \frac{1}{2} g t^2[/tex]
and so
[tex]t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 90m}{9.81 m/s^2} }=4.28 s [/tex]
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
[tex]v_x= \frac{S_x}{t} = \frac{180 m}{4.28 s} =42 m/s[/tex]
