Answer to Question #134058 in Mechanics | Relativity for David P. Costello

Question

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 5.93 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

Givens;

v=0

a=-9.8m/s^2

$v=v_o+at$ but v=0,

$0=v_o-gt_1$

$t_1=\frac{v_o}{g}$

therefore $v_0=t_1g$ . Also $t_1$ can be obtained by

$v^{2}=v_0^{2}-2gh$ , substituting $v_0$ in the equation

$v^2=(t_1g)^2-2gh$

$t_1=\sqrt\frac{2h}{g}$

time taken by ball for downward motion

$v _0=0$

$y=v_0t+\frac{1}{2}at_2^{2}$

$-h=0-\frac{1}{2}gt^2$

making $t_2$ subject of the formula

$t_2=\sqrt\frac{2h}{g}$

total airtime of the ball is therefore,

T= $t_1+t_2$

T$=2\sqrt\frac{2h}{g}$

$2\times \sqrt\frac{2h}{9.8}=5.93$

$h=\frac{(\frac{5.93}{2})^2\times9.8}{2}=43.077m$

Therefore h=43.077m