Question

a baseball is hit straight up into the air. If the initial velocity was 20 m/s, how high will the ball go?How long will it be until the catcher catches the ball at the same height it was hit? How fast is it going when catcher catches it?

Answer 1

1) The motion of the ball is an uniformly accelerated motion, with constant acceleration equal to $g=-10 m/s^2$ (the negative sign means it is directed towards the ground).

For an uniformly accelerated motion, we can use the following relationship:
$2gh=v_f^2-v_i^2$
where h is the maximum height reached by the ball, $v_i = 20 m/s$ is its initial velocity and $v_f$ the velocity of the ball when it reaches the maximum height. But $v_f=0$ (when the ball reaches the maximum height, it stops before going down, so its velocity at that moment is zero), so we can use the relationship to  calculate h, the maximum height:
$h=- \frac{v_i^2}{2g} =- \frac{20 m/s)^2}{2 \cdot (-10 m/s^2)} =20 m$

2) We can find the time the ball takes to return to the ground by requiring that the space covered by the ball returns to zero: 
$S(t)=0$
where for an uniformly accelerated motion,
$S(t)=v_i t + \frac{1}{2} gt^2 =0$
By solving this, we have two solutions: one is t=0, which corresponds to the moment the player hits the ball, the second one is
$t=- \frac{2 v_i}{g}=- \frac{2 \cdot 20 m/s}{-10 m/s^2}=4 s$
so, the ball returns to the ground after 4 s.

3) The velocity of the ball when it returns to the ground is given by:
$v(t) = v_i +gt=20 m/s + (-10 m/s^2)(4 s)=-20 m/s$
where the negative sign means the ball is going downwards.