Answers:
a) 3.511 s
b) 15.1062 m
c) 86.281 m
Explanation:
Assuming the initial velocity of the ball is [tex]V_{o}=30 m/s[/tex] and the angle [tex]\theta=30\°[/tex], we have the following data:
[tex]d=40 yards=36.58 m[/tex] the distance between the football player and the the goal post
[tex]H=10 ft=3.084 m[/tex] the height of the goal post
Now, this can be solved with the following equations related to parabolic motion:
[tex]t_{flight}=\frac{2V_{o} sin \theta}{g}[/tex] (1)
[tex]y_{max}=y_{o}+V_{o}sin \theta \frac{t_{flight}}{2}-\frac{g}{2}(\frac{t_{flight}}{2})^{2}[/tex] (2)
[tex]x=V_{o}cos\theta t_{flight}[/tex] (3)
Where:
[tex]t_{flight}[/tex] is the time of flight
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]y_{max}[/tex] is the maximum height, when the time is half the time of flight [tex]t_{flight}[/tex]
[tex]y_{o}=0 m[/tex] is the initial height
[tex]x[/tex] is the horizontal range
Knowing this, let's begin with the answers:
We will use equation (1):
[tex]t_{flight}=\frac{2(30 m/s) sin(30\°)}{9.8 m/s^{2}}[/tex] (4)
[tex]t_{flight}=3.511 s[/tex] (5)
In this case, we have to use equation (2) and substitute the [tex]t_{flight}[/tex] calculated in (5):
[tex]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}[/tex] (6)
[tex]y_{max}=15.1062 m[/tex] (7)
Let's use equation (3):
[tex]x=(30m/s)cos(30\°) (3.511 s)[/tex] (8)
[tex]x=86.281 m[/tex] (9)
Since the maximum height and the horizontal range of the ball are greater than [tex]d[/tex] and [tex]H[/tex], we can say the kicker is able to make the field goal.
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 [tex]\alpha[/tex]= [tex]45^{o}[/tex](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= [tex]\frac{2Vi sin\alpha }{g}[/tex] = [tex]\frac{2(40) sin 45^{o} }{9.8}[/tex]
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
