Respuesta :
Answer:
The kick is good. The ball clears the crossbar by nearly 15 feet.
Step-by-step explanation:
Notice that the ball is a projectile, its movement is a parabolla.
We know that [tex]v_{0}=70 \ ft/sec[/tex] and the distance Sam attempts is 42 yards, where the crossbar is 10 feet above.
We must find the maximum height and the maximum horizontal distance, to compare with the position of the crossbar, that will determine if the kick was good enough or not.
[tex]x_{max}=v_{x}t_{total}[/tex]
Notice that [tex]v_{x} =v_{x_{0} }[/tex], because gravity doesn't affect horizontal coordinates.
Also, we can find the vertical initial velocity and the horizontal initial velocity
[tex]v_{0_{y} }=v_{0} sin 50\°=70 (sin50\°) \approx 52.62 \ ft/sec[/tex]
[tex]v_{0_{x} }= v_{0} cos50\°=70(cos50\°) \approx 45 \ ft/sec[/tex]
Now, we can find the maxium height, where [tex]v_{f}=0[/tex], because the ball stops at the maxium height point.
[tex]v_{f}^{2} =v_{0}^{2}+2gh\\0=v_{0}^{2}+2gh_{max} \\-v_{0}^{2}=2gh_{max}\\h_{max}=\frac{-v_{0}^{2}}{2g}= \frac{-(52.62)^{2} }{2(-32.2)} \\h_{max} \approx 43 \ ft[/tex]
Apparently, the ball has enough height to pass above the crossbar.
Now, we find the total time of the movement to find the maximum horizontal distance of the ball.
[tex]v_{f} =v_{0}+gt \\0=52.62-32.2t\\t=\frac{-52.62}{-32.2} \approx 1.63 \ sec[/tex]
The total time is
[tex]t_{total}=2t=2(1.63)=3.26 \ sec[/tex]
Then, we find the maximum horizontal distance
[tex]x_{max}=v_{x}t_{total}=45(3.26)=146.7 \ ft[/tex]
This means the ball will travel 146.7 feet horizontally.
Now, we need to know by how much distance the ball passed over the crossbar. We know that the crossbar is 42 yards away, which is equal to 126 feet.
[tex]t_{crossbar}=\frac{126}{45}= 2.8 \ sec[/tex]
But, we only need the time of the falling part of the movement
[tex]t=2.8-1.63=1.17 \ sec[/tex]
Now, let's find the vertical position
[tex]y=v_{0}t+\frac{1}{2}gt^{2}\\ y=\frac{1}{2}(32.2)(1.17)^{2} \approx 22.04 \ ft[/tex]
Then, we subtract
[tex]43-22.04 = 20.96 \ ft[/tex]
And, [tex]20.96 - 10 = 10.96 \ ft[/tex]
(We used a lot of approximation, that's why gives a highe number than it should)
Therefore, the kick is good. The ball clears the crossbar by nearly 15 feet.
Answer:
The correct option is;
The kick is good! The ball clears the crossbar by nearly 15 feet.
Step-by-step explanation:
Here we have;
Velocity of ball = 70 ft/s
Angle of motion = 50°
Therefore;
Vertical component of velocity, = v×sinθ = v×sin×50 = 53.62 ft/s
= u - g·t
At maximum height = 0, therefore;
u = g·t and
t = u/g
Where:
u = Initial vertical velocity = 53.62 ft/s
g = Acceleration due to gravity = 32.1740 ft/s²
∴ t = 53.62/32.1740 = 1.66666 s
Total time of flight = 2 × 1.66666 = 3.33332 s
Max height is given by the following relation;
v² = u² - 2·g·s
v = 0 ft/s at maximum height, therefore;
u² = 2·g·s
53.62² = 2×32.1740×s
s = 53.62² ÷(2×32.1740) = 44.69 ft
Horizontal velocity = 70×cos(50) = 44.995 ft/s
Hence the time it takes the ball to reach 42 yards is given by the following relation;
42 yards = 126 ft
Height of ball at 2.8 s is given by;
Direction of ball at 2.8 s = downwards
Hence time after peak = 2.8 - 1.666 = 1.1336429 s
Position at 2.8 s is given by
s = u ·t + 1/2·g·t²
u = 0 ft/s as ball is on second half of flight
∴ s = 1/2·g·t² = 1/2×32.1740×1.1336429²= 20.674 ft below maximum height
∴ Height above ground = Maximum height - 20.674 ft = 44.69 - 20.674
Height above ground = 24.01 ft
Hence the ball clears the crossbar by 24.01 - 10 = 14.01 ft
The best option is therefore;
The kick is good! The ball clears the crossbar by nearly 15 feet.
