Question

During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 6.41 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 m

Answer 1

Given

Time of flight is$t=6.41 s$

The expression for the time of flight is

$t=\frac{2u}{g}$

The initial velocity is

$u=\frac{gt}{2}=\frac{(9.80)(6.41)}{2}=$ 31.4 m/s

The maximum height is

$h_{max}=\frac{u^2}{2g}=\frac{(31.4)^2}{2(9.80)}=50 m$