Question

During a baseball game, a batter hits a popup to a fielder 93 m away. The acceleration of gravity is 9.8 m/s^2 . If the ball remains in the air for 5.4 s, how high does it rise? Answer in units of m.

Answer 1

$\large T=\frac{2usin\theta}{g}$

$\large 5.4s=\frac{2usin\theta}{9.8m/s^2}$

$\large usin\theta=\frac{5.4s\times 9.8m/s^2}{2}=26.46m/s$

$\large h=\frac{u^2sin^2\theta}{2g}=\frac{(usin\theta)^2}{2g}$

$\large h=\frac{(usin\theta)^2}{2g}=\frac{(26.46)^2}{2\times 9.8}$

$\large h=35.72m$