1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
[tex]x(t)=v_0t[/tex]
[tex]y(t)=h- \frac{1}{2}gt^2 [/tex]
where the motion on the x-axis is a uniform motion with constant speed [tex]v_0=18.1 m/s[/tex], while the motion on the y-axis is a uniformly accelerated motion with constant acceleration [tex]g=9.81 m/s^2[/tex], and initial height [tex]h=1.85 m[/tex].
First of all, we can find the time t at which the ball reaches the ground by requiring [tex]y(t)=0[/tex]:
[tex]0=h- \frac{1}{2}gt^2 [/tex]
[tex]t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }= 0.61 s[/tex]
and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
[tex]x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m[/tex]
2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
[tex]v_x=18.1 m/s[/tex]
On the y-axis, the velocity is given by
[tex]v_y(t)=gt[/tex]
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
[tex]v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s[/tex]
And the speed of the ball is the magnitude of the resultant of the two components:
[tex]v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s [/tex]
