A punter in a football game kicks a ball from the goal 60° from the horizontal at 25 m/s. What is the hang time afthe punt, and how far down field does the ball land?

Question

Charlotte Aaron · Accepted Answer

#(a) " " 4.42"s"#

#(b)" " 55.25"m"# #(a)# Using the vertical component of the velocity, we can calculate the time needed to reach the highest point: #v=u+at# #:.0=25sin60-9.8t# #:.t=61.65/9.8=2.21"s"# Assuming fair play, it follows that the return to Earth will take the same amount of time. #:. # total time in flight #=2xx2.21=4.42"s"# #(b)# Since the velocity's horizontal component is constant, we obtain: Distance is equal to speed times time. #=25cos60xx4.42=55.25"m"#

Charlotte Aaron · Answer

undefined To find the hang time of the punt, we can use the following equation:

$$ hang\ time = \frac{2 \cdot v_0 \cdot sin(\theta)}{g} $$

where:
- $ v_0 $ is the initial velocity (25 m/s)
- $ \theta $ is the angle of projection (60°)
- $ g $ is the acceleration due to gravity (approximately 9.8 m/s²)

Substituting the given values:

$$ hang\ time = \frac{2 \cdot 25 \cdot sin(60°)}{9.8} $$
$$ hang\ time \approx \frac{2 \cdot 25 \cdot \frac{\sqrt{3}}{2}}{9.8} $$
$$ hang\ time \approx \frac{25 \cdot \sqrt{3}}{4.9} $$
$$ hang\ time \approx \frac{25 \cdot 1.732}{4.9} $$
$$ hang\ time \approx \frac{43.3}{4.9} $$
$$ hang\ time \approx 8.85 $$ seconds

To find the horizontal distance the ball travels, we can use the following equation:

$$ distance = v_0 \cdot cos(\theta) \cdot hang\ time $$

Substituting the given values:

$$ distance = 25 \cdot cos(60°) \cdot 8.85 $$
$$ distance = 25 \cdot \frac{1}{2} \cdot 8.85 $$
$$ distance = 112.5 $$ meters

Therefore, the hang time of the punt is approximately 8.85 seconds, and the ball lands approximately 112.5 meters downfield.