Question

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where it hits the bat does it rise? Assume when it hits the ground it hits at exactly the level of the bat. The acceleration of gravity is 9.8 m/s 2 . Answer in units of

Answer 1

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

$d=vt+\frac{1}{2}at^2$

Now by substituting values in, we get

$d=(0\frac{m}{s})*(2.01s)+\frac{1}{2}(9.8\frac{m}{s^2})(2.01)^2$

d = 19.796m