Question

Ayden and Steven are playing catch with a football. Ayden throws the football at a velocity of 15 m/s with an angle of 60° above the horizontal. What is the hang time of the football? What is the maximum height of the football? About how far away should Steven stand so he can catch the football?

Answer 1

1) 1.33 s

2) 8.6 m

3) 19.9 m

Explanation:

1)

The motion of the football is a projectile motion, which consists of two independent motions:

- A uniform motion (=constant velocity) along the horizontal direction

- A uniformly accelerated motion (=constant acceleration) along the vertical direction

The hang time of the football is the time it takes for the football to reach its maximum height. It is given by:

$t=\frac{u_y}{g}$

where

$u_y = u sin \theta$ is the initial vertical velocity of the ball, where

u = 15 m/s is the initial velocity

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

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

Substituting, we find the hang time:

$t=\frac{u sin \theta}{g}=\frac{(15)(sin 60^{\circ})}{9.8}=1.33 s$

2)

The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where:

s is the vertical displacement

$u_y = u sin \theta$ is the initial vertical velocity

t is the time

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

In this part, we want to find the maximum height, so the height reached by the ball when the time is

t = 1.33 s

Therefore, by substituting the values in the equation, we can find the maximum height:

$s=usin \theta t-\frac{1}{2}gt^2=(15)(sin 60^{\circ})(1.33)-\frac{1}{2}(9.8)(1.33)^2=8.6 m$

3)

Here we want to find the horizontal range covered by the ball during its flight.

The horizontal range for a projectile is given by the equation

$d=\frac{u^2 sin(2\theta)}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration due to gravity

For the ball in this problem we have:

u = 15 m/s

$\theta=60^{\circ}$

$g=9.8 m/s^2$

Substituting into the equation, we  find the horizontal distance covered by the ball:

$d=\frac{(15)^2sin(2\cdot 60^{\circ})}{9.8}=19.9 m$