The motion of the ball is a composition of the two motions:
- on the horizontal axis, it is a uniform motion with constant speed [tex]v_x = 15 m/s[/tex]
- on the vertical axis, it is a uniformly accelerated motion with initial speed [tex]v_y =0[/tex] and constant acceleration [tex]g=9.81 m/s^2[/tex] toward the ground
The laws of motion on the two directions are:
[tex]x(t) = v_x t [/tex]
[tex]y(t) = h - \frac{1}{2}gt^2 [/tex]
where h=78.4 m is the initial altitude of the ball.
We want to find how far the ball travels on the x-axis. In order to do that, we must find the time t at which the ball reaches the ground, i.e. the time t at which y(t)=0:
[tex]0=h- \frac{1}{2}gt^2 [/tex]
[tex]t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2 \cdot 78.4 m}{9.81 m/s^2} }=4.0 s [/tex]
And so now we can find the distance x(t) covered by the ball during this time:
[tex]x(t) = v_x t = (15 m/s)(4.0 s)=60 m[/tex]
