a place kicker kicks a football with a velocity of 20.0 minutes per hour and at an angle at 53 degrees

a.how long is the ball in the air
b.how far away does it land
c.how high does it travel

20 minutes per hour makes no sense. So let's assume:

Vo = 20m/s[53o].
Xo = 20*Cos53 = 12.0 m/s.
Yo = 20*sin53 = 16.0 m/s.

a. Y = Yo + g*Tr = 0, 16 - 9.8*Tr = 0, -9.8Tr = -16, Tr = 1.63 s. = Rise time.

Tf = Tr = 1.63 s. = Fall time.

T = Tr+Tf = 1.63 + 1.63 = 3.27 s. = Time in air.

b. d = Xo*T = 12 * 3.27 = 39.2 m.

c. Y^2 = Yo^2 + 2g*h. Y = 0, Yo = 16 m/s, g = -9.8 m/s^2, h = ?.

a. Well, it depends. Is the ball an introvert or an extrovert? If it's an introvert, it might want to come back to the ground quickly. But if it's an extrovert, maybe it'll enjoy flying through the air a bit longer. Let's assume it's a party-loving extrovert, so it'll stay in the air for a while.

b. Now, I'm no mathematician, but I do know that the football is going to act like your ex and go as far away from you as possible. So, get ready for some distance!

c. Ah, the height. You know, footballs have a remarkable ability to dream big and reach for the sky. I wouldn't be surprised if this ball tries to touch the clouds.

To solve this problem, we will use the following equations:

1. Vertical motion equation: h = vi * t * sin(theta) - (1/2) * g * t^2
2. Horizontal motion equation: d = vi * t * cos(theta)

Given:
Initial velocity (vi) = 20.0 minutes per hour
Launch angle (theta) = 53 degrees
Gravitational acceleration (g) = 9.8 m/s^2

a. How long is the ball in the air?

To find the time of flight (t), we need to determine when the vertical displacement (h) becomes zero. We can use the above equation and solve for t.

Set h = 0:
0 = vi * t * sin(theta) - (1/2) * g * t^2

Rearranging, we have:
(1/2) * g * t^2 = vi * t * sin(theta)

Dividing both sides by t:
(1/2) * g * t = vi * sin(theta)

Simplifying further:
t = (2 * vi * sin(theta)) / g

Plugging in the values:
t = (2 * 20.0 minutes per hour * sin(53 degrees)) / 9.8 m/s^2

Converting minutes per hour to meters per second:
t = (2 * 20.0 * 0.267949) / 9.8
t ≈ 1.15 seconds

Therefore, the ball is in the air for approximately 1.15 seconds.

b. How far away does it land?

To find the horizontal distance (d), we can use the horizontal motion equation mentioned above.

d = vi * t * cos(theta)
= 20.0 minutes per hour * 1.15 seconds * cos(53 degrees)

Again, converting minutes per hour to meters per second:
d = 20.0 * 0.267949 * 1.15 * cos(53 degrees)

d ≈ 12.63 meters

Therefore, the ball lands approximately 12.63 meters away.

c. How high does it travel?

To find the maximum height (h_max) reached by the ball, we can substitute the time of flight (t) into the vertical motion equation and solve for h.

h = vi * t * sin(theta) - (1/2) * g * t^2
= 20.0 * 0.267949 * 1.15 - (1/2) * 9.8 * 1.15^2

h ≈ 2.28 meters

Therefore, the ball travels approximately 2.28 meters high.

To answer these questions, we need to use basic kinematic equations for projectile motion. Here's how to calculate each part:

a. To find how long the ball is in the air, we can substitute the given initial velocity (20.0 minutes per hour) and angle (53 degrees) into the equation:

time = (2 * initial velocity * sin(angle)) / gravitational acceleration

In this equation, gravitational acceleration is approximately 9.8 m/s². However, there seems to be an error in the given velocity unit (minutes per hour), as velocity should be given in a unit like meters per second. If you provide the velocity in a valid unit, we can continue the calculation.

b. To determine how far away the ball lands, we can use the equation for horizontal distance:

horizontal distance = initial velocity * time * cos(angle)

By substituting the values for the initial velocity, time (calculated in part a), and angle, we can find the answer.

c. To calculate the maximum height the ball reaches, we use the equation for vertical displacement:

vertical displacement = (initial velocity² * sin²(angle)) / (2 * gravitational acceleration)

By substituting the values for the initial velocity and angle, we can find the answer.

If you can provide the velocity in a valid unit, I can help you calculate the answers for parts a, b, and c.