Question

A kicker on a sportsball team tries to kick a sportsball so that it stays in the air for a long "hang time." If the ball is kicked with an initial velocity of 24.8 m/s at an angle of 60.0° above the ground, what is the "hang time"?

Answer 1

Answer:

4.39 s

Explanation:

The motion of the ball is a projectile motion, whose time of flight (the time it remains in the air) is given by

$t=2\frac{u_y}{g}$ (1)

where

$u_y$ is the initial vertical velocity of the projectile

$g=9.8 m/s^2$ is the acceleration of gravity

The initial vertical velocity of the ball is given by

$u_y = u sin \theta$

where

u = 24.8 m/s is the initial speed

$\theta=60^{\circ}$ is the angle of projection

Substituting,

$u_y = (24.8)(sin 60)=21.5 m/s$

And now by substituting into (1), we find the time of flight (the "hang time"):

$t=\frac{2(21.5)}{9.8}=4.39 s$