(a) The motion on the vertical axis is an uniformly accelerated motion, with acceleration [tex]g=9.81 m/s^2[/tex] and initial vertical velocity [tex]v_y = 13 m/s[/tex], so the vertical position of the ball at time t is:
[tex]y(t) = v_y t - \frac{1}{2}gt^2 [/tex]
where we put a minus sign in front of g because the gravitational acceleration points toward the ground. The time when the ball hits the ground is the time t at which the vertical position is zero:
[tex]y(t) = v_y t - \frac{1}{2}gt^2 =0 [/tex]
which gives two solutions:
[tex]t=0 s[/tex], the initial moment when the ball is thrown,
[tex]t= \frac{2v_y}{g}= \frac{2 \cdot 13 m/s}{9.81 m/s^2}=2.65 s [/tex], the moment when the ball hits the ground.
The vertical velocity at time t is given by
[tex]v_y(t) = v_y - gt[/tex]
and by substituting [tex]t=2.65 s[/tex], we find the vertical velocity of the ball when it hits the ground
[tex]v_y(2.65 s) = 13 m/s - (9.81 m/s^2)(2.65 s)=-13.00 s[/tex]
So, it is equal to the initial vertical velocity, but in the opposite direction (toward the ground, due to the negative sign).
To find the speed of the ball when it hits the ground, we should find the resultant of the horizontal velocity (which has not changed, because the motion on the horizontal axis is a uniform motion) and of the vertical velocity:
[tex]v= \sqrt{v_x^2 + v_y^2}= \sqrt{(22 m/s)^2+(-13 m/s)^2} =25.6 m/s [/tex]
and this is the speed of the ball when it hits the ground.
(b) we already found the answer at point (a): the time it takes the ball to reach the ground is 2.65 s, so this is the time the ball remains in the air.
(c) The maximum height reached by the ball is the vertical position y(t) at which the vertical velocity is zero:
[tex]v_y (t) = 0[/tex]
Therefore,
[tex]v_y - gt = 0[/tex]
[tex]t= \frac{v_y}{g}= \frac{13 m/s}{9.81 m/s^2}=1.33 s [/tex]
this is the time at which the ball reaches the maximum height, and if we use this value in the formula of y(t), we find the value of the maximum height:
[tex]h=y(1.33 s)=v_y t - \frac{1}{2}gt^2=(13 m/s)(1.33 s)- \frac{1}{2}(9.81 m/s^2)(1.33s)^2 =8.6 m [/tex]
