Question

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.89s, how high does it rise?

Answer 1

Let’s first find the initial velocity of the ball from the kinematic equation:
v=v_0-gt_rise,
here, v=0 is the velocity of the ball at the maximum height, v_0 is the initial velocity of the ball, g=9.8 m⁄s^2 is the acceleration due to gravity, t_rise=t⁄2 is the time that the ball needs to attain the maximum height and t=4.89 s is the time of flight.
Then, we get:
0=v_0-g t/2,
v_0=g t/2=9.8 m/s^2 ∙(4.89 s)/2=24 m/s.
Finally, we can find the maximum height that the ball can attain from another kinematic equation:
y_max=v_0 t_rise-1/2 gt_rise^2=24 m/s∙2.44 s-1/2∙9.8 m/s^2 ∙(2.44 s)^2=29.4 m