So A rookie quarterback throws a football with an initial upward velocity component of 16.4 m/s and a horizontal velocity component of 19.8 m/s . Ignore air resistance. a) How much time is required for the football to reach the highest point of the trajectory? b) How high is this point? c) How much time (after it is thrown) is required for the football to return to its original level? d) How does this compare with the time calculated in part (a).

Express your answer using three significant figures. e) How far has it traveled horizontally during this time?

Xo = 19.8 m/s.

Yo = 16.4 m/s.

a. Y = Yo + g*Tr = 0, Tr = -Yo/g = -16.4/-9.8 = 1.67 s. = Rise time.

b. h = Yo*Tr + 0.5g*Tr^2.
g = -9.8 m/s^2.

c. T = Tr+Tf, Tf = Tr = 1.67 = Fall time.
T = Tr+Tr = 1.67 + 1.67 =

d. T = 2Tr.

e. d = Xo*T.

To solve this problem, we can break it down into several steps. Let's go through each part step by step:

a) How much time is required for the football to reach the highest point of the trajectory?
Since we are ignoring air resistance, the only force acting on the football in the vertical direction is gravity. The initial upward velocity will gradually decrease until it reaches zero at the highest point. Therefore, we can use the following equation to determine the time (t) it takes for the football to reach the highest point:

v_final = v_initial + (acceleration * t)

Here, v_initial is the initial vertical velocity, v_final is the final vertical velocity (which is zero at the highest point), and acceleration is the acceleration due to gravity (-9.8 m/s^2).

0 m/s = 16.4 m/s + (-9.8 m/s^2 * t)

Simplifying the equation, we have:

9.8 m/s^2 * t = 16.4 m/s

t = 16.4 m/s / 9.8 m/s^2

Calculating t, we find:

t ≈ 1.67 s

So it takes approximately 1.67 seconds for the football to reach the highest point of its trajectory.

b) How high is this point?
To find the height at the highest point, we can use the equation:

height = v_initial * t + (1/2) * acceleration * t^2

Here, v_initial is the initial vertical velocity, t is the time calculated in part (a), and acceleration is the acceleration due to gravity (-9.8 m/s^2).

height = 16.4 m/s * 1.67 s + (1/2) * (-9.8 m/s^2) * (1.67 s)^2

Calculating the height, we find:

height ≈ 13.73 m

So the highest point of the football's trajectory is approximately 13.73 meters.

c) How much time (after it is thrown) is required for the football to return to its original level?
Since the football stops momentarily at the highest point, the time it takes to return to its original level would be double the time calculated in part (a).

time_return = 2 * t

Substituting the value of t, we have:

time_return = 2 * 1.67 s

Calculating the time_return, we find:

time_return ≈ 3.34 s

So it takes approximately 3.34 seconds for the football to return to its original level.

d) How does this compare with the time calculated in part (a)?
The time calculated in part (a) was approximately 1.67 seconds, while the time calculated in part (c) was approximately 3.34 seconds. Thus, the time it takes for the football to return to its original level is twice the time it takes to reach the highest point.

e) How far has it traveled horizontally during this time?
To determine the horizontal distance traveled during the given time, we can use the formula:

distance = velocity * time

Here, velocity is the initial horizontal velocity component, and time is the time calculated in part (c).

distance = 19.8 m/s * 3.34 s

Calculating the distance, we find:

distance ≈ 66.13 m

So the football has traveled approximately 66.13 meters horizontally during this time.

To answer these questions, we can use the principles of projectile motion. Let's break down each part of the question and explain how to get the answers:

a) How much time is required for the football to reach the highest point of the trajectory?

The vertical motion (upward and downward) of the football is affected by gravity. The initial upward velocity component is 16.4 m/s. At the highest point of the trajectory, the vertical velocity will be zero. To find the time it takes to reach this point, we can use the formula:

v = u + at

where:
- v is the final velocity (which is 0 m/s at the highest point)
- u is the initial velocity (16.4 m/s)
- a is the acceleration due to gravity (-9.8 m/s², as it acts in the opposite direction of the initial velocity)
- t is the time.

Rearranging the formula to solve for time gives us:

t = (v - u) / a

Substituting the given values:

t = (0 - 16.4) / -9.8

Calculating this gives us the time required for the football to reach the highest point.

b) How high is this point?

To determine the height at the highest point, we can use the formula for the vertical displacement:

s = ut + (1/2)at²

where:
- s is the displacement (height in this case)
- u is the initial velocity (16.4 m/s)
- t is the time (which we calculated in part a)
- a is the acceleration due to gravity (-9.8 m/s²).

Substituting the given values and calculating will give us the height at the highest point.

c) How much time (after it is thrown) is required for the football to return to its original level?

When the football returns to its original level, its vertical displacement will be zero. Using the same formula as part b, we can set s = 0 and solve for t. This will give us the time it takes for the football to return to its original level.

d) How does this compare with the time calculated in part a?

Compare the values obtained in parts a and c. This will give us the comparison between the time required to reach the highest point and the time required to return to the original level.

e) How far has it traveled horizontally during this time?

The horizontal motion of the football is not affected by gravity since there's no horizontal acceleration. Hence, it will continue to move horizontally with a constant velocity of 19.8 m/s throughout its flight. Multiplying the horizontal velocity by the time calculated in part c will give us the distance traveled horizontally.

By following these steps, you can calculate the answers to each part of the question.