Question

what would be the time of flight, maxiumum height and horizontal range of the football was kicked from the floor? if a field goal attempt was made from 40 yards (36.58 m) and the goal post is 10 feet tall (3.084 m) will the kicker make the field goal?

Answer 1

Answer:

time of flight T= 5.77sec

maxiumum height h max= 66.78m

horizontal range  S=50.91m

Explanation:

Assume that the footaball moves with the initial velocity= Vo=40m/s

and angle $\alpha$= $45^{o}$(for maximum range)

Given that,

the distance from place of attempting goal to goal post =d= 36.58m

height of goal post=3.084m

For projectile motion, time of flight T is,

T= $\frac{2Vi sin\alpha }{g}$ = $\frac{2(40) sin 45^{o} }{9.8}$

T= 5.77sec

According to second equation of motion

S=Vot+ 1/2 gt^2

S= h max,   Vo= vertical component of velocity= 40sin 45=28.28

t=average time= (time of flight +0 )/2= 3.6/2=1.8sec

h max= 28.28*1.8+ 1/2(9.8)(1.8)^2

h max= 66.78m

For horizontal range,

S= v*t

v= horizontal component of velocity= Vocos45=40cos 45 = 28.28

t=average time= (time of flight +0 )/2= 3.6/2=1.8sec

S= 28.28*1.8

S=50.91m

Answer 2

Answers:

a) 3.511 s

b) 15.1062 m

c) 86.281 m

Explanation:

Assuming the initial velocity of the ball is $V_{o}=30 m/s$ and the angle  $\theta=30\°$, we have the following data:

$d=40 yards=36.58 m$ the distance between the football player and the the goal post

$H=10 ft=3.084 m$ the height of the goal post

Now, this can be solved with the following equations related to parabolic motion:

$t_{flight}=\frac{2V_{o} sin \theta}{g}$ (1)

$y_{max}=y_{o}+V_{o}sin \theta \frac{t_{flight}}{2}-\frac{g}{2}(\frac{t_{flight}}{2})^{2}$ (2)

$x=V_{o}cos\theta t_{flight}$ (3)

Where:

$t_{flight}$ is the time of flight

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

$y_{max}$ is the maximum height, when the time is half the time of flight $t_{flight}$

$y_{o}=0 m$ is the initial height

$x$ is the horizontal range

Knowing this, let's begin with the answers:

a) Time of flight

We will use equation (1):

$t_{flight}=\frac{2(30 m/s) sin(30\°)}{9.8 m/s^{2}}$ (4)

$t_{flight}=3.511 s$ (5)

b) Maximum height

In this case, we have to use equation (2) and substitute the $t_{flight}$ calculated  in (5):

$y_{max}=(30 m/s) sin(30\°) \frac{3.511 s}{2}-\frac{9.8 m/s^{2}}{2}(\frac{3.511 s}{2})^{2}$ (6)

$y_{max}=15.1062 m$ (7)

c) Horizontal range

Let's use equation (3):

$x=(30m/s)cos(30\°) (3.511 s)$ (8)

$x=86.281 m$ (9)

Since the maximum height and the horizontal range of the ball are greater than $d$ and $H$, we can say the kicker is able to make the field goal.