Question

A baseball pitcher throws a ball vertically straight upward and catches the ball 4.9 s later.

(calculators permitted in this question)

(a) With what velocity did the ball leave the pitcher's hand?

v?i =  m/s

(b) What was the maximum height reached by the ball?

h =  m

Answer 1

Acceleration due to gravity  $g=-9.81\,m/s^2$ (since acceleration is downwards)

Initial velocity $v_i$

Time of flight $T=4.9\,s$

(a)

Using formula for displacement $s=v_it+\frac{1}{2}gt^2$

Final displacement is zero since the ball reached the point of projection.

$0=v_i*4.9-\frac{1}{2}*9.81*4.9^2$

$v_i=\frac{1}{2}*9.81*4.9=24.03\,m/s$

(b)

At maximum height velocity of ball is zero.

$v_f^2-v_i^2=2gh$

$0-\left(24.0345 \right )^2=-2*9.81*h$

Maximum height $h=\frac{-(24.0345)^2}{-2*9.81}=29.44\,m$