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:
[tex]t=\frac{u_y}{g}[/tex]
where
[tex]u_y = u sin \theta[/tex] is the initial vertical velocity of the ball, where
u = 15 m/s is the initial velocity
[tex]\theta=60^{\circ}[/tex] is the angle of projection of the ball
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Substituting, we find the hang time:
[tex]t=\frac{u sin \theta}{g}=\frac{(15)(sin 60^{\circ})}{9.8}=1.33 s[/tex]
2)
The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:
[tex]s=u_y t - \frac{1}{2}gt^2[/tex]
where:
s is the vertical displacement
[tex]u_y = u sin \theta[/tex] is the initial vertical velocity
t is the time
[tex]g=9.8 m/s^2[/tex] 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:
[tex]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[/tex]
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
[tex]d=\frac{u^2 sin(2\theta)}{g}[/tex]
where
u is the initial velocity
[tex]\theta[/tex] is the angle of projection
g is the acceleration due to gravity
For the ball in this problem we have:
u = 15 m/s
[tex]\theta=60^{\circ}[/tex]
[tex]g=9.8 m/s^2[/tex]
Substituting into the equation, we find the horizontal distance covered by the ball:
[tex]d=\frac{(15)^2sin(2\cdot 60^{\circ})}{9.8}=19.9 m[/tex]
