Jake attempts a 36-yard field goal in a football game. For his attempt to be a success, the football needs to pass through the u
1 answer:
Question:
1) The kick is good! The ball passes about 10 feet above the crossbar.
2) The kick is good! The ball clears the crossbar by approximately 14 feet.
3) The kick is good! The ball clears the crossbar by about 4 feet.
4) The kick is not successful. The ball is 2 feet too low.
Answer:
The correct option is;
3) The kick is good! The ball clears the crossbar by about 4 feet.
Step-by-step explanation:
Here we have
Horizontal distance from goal, x = 36 yards = 108 ft
Vertical distance of goal, y = 10 ft
Angle of elevation of ball = 34°
Initial velocity of ball, v₀ = 68 ft/s
∴ Vertical component of velocity = = v₀×sin(θ₀)×t - g×t
Horizontal component of velocity = vₓ = 68×cos(34) = 56.375 ft/s
The equation for projectile motion is as follows
x = x₀ + v₀×cos(θ₀)×t
y = y₀ + v₀×sin(θ₀)×t - 1/2×g×t²
Where:
x₀ = Initial horizontal displacement of the ball = 0
y₀ = Initial vertical displacement of the ball = 0
t = Time of flight of the ball
Therefore;
108 = 0 + 68×cos(34)×t which gives;
t = 108/56.375 = 1.916 seconds
Hence, y = y₀ + v₀×sin(θ₀)×t - 1/2×g×t² gives;
y = 0 + 68×sin(34)×1.916 - 1/2×32.2×1.916²
y = 72.85 - 59.09 = 13.76 ft, hence the vertical location of the ball at 36 yard is 13.76 ft where the crossbar is at 10 ft hence the ball clears the crossbar by 13.76 ft - 10 ft = 3.76 ft ≈ 4 ft
Therefore, the kick is good! The ball clears the crossbar by about 4 feet.
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Answer:
a) The 99% confidence interval would be given by (589.588;731.038)
b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The data is:
661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706
Part a
Compute the sample mean and sample standard deviation.
In order to calculate the mean and the sample deviation we need to have on mind the following formulas:
=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)
On this case the average is
=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)
The sample standard deviation obtained was s=95.898
Find the critical value t* Use the formula for a CI to find upper and lower endpoints
In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by and
. The degrees of freedom are given by:
We can find the critical values in excel using the following formulas:
"=T.INV(0.005,15)" for
"=T.INV(1-0.005,15)" for
The confidence interval for the mean is given by the following formula:
And if we find the limits we got:
[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]
So the 99% confidence interval would be given by (589.588;731.038)
Part b
If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.