Aplace kicker kicks a football at an angle of 42 degrees above the ground with a speed of 22 m/s. a) find time to max. height? b) find time to the ground? c) find max. height of the ball? d) find the horizontal displacement of the kickoff?

A) The time to the maximum height is 1.50 seconds

B) The time to the ground is 3.0 seconds

C) The maximum height is 11.06 meters

D) The horizontal displacement of the kickoff

Explanation:

A place kicker kicks a football at an angle of 42 degrees above the

ground with a speed of 22 m/s

→ u = 22 m/s and Ф = 42°

where u is the initial velocity and Ф is its direction

At first lets find the initial components of velocity

The initial horizontal component is u cosФ

→ The initial horizontal component is 22 cos(42)

The initial vertical component is u sinФ

→ The initial vertical component is 22 sin(42)

A)

At the maximum height the vertical component of velocity is zero

The vertical component of velocity after t seconds is

→ = u sinФ + g t

Where g is the acceleration of gravity and t is the time

→ g = -9.8 m/s² , = 0

→ 0 = 22 sin(42) - 9.8 t

Add 9.8 t for both sides

→ 9.8 t = 22 sin(42)

Divide both sides by 9.8

→ t = 1.50 seconds

The time to the maximum height is 1.50 seconds

B)

At the ground the vertical displacement y = 0

So The ball will fall from the maximum height again

Then it will take the same time for the maximum height

The time of the flight of the ball from the ground to the ground again is

double the time from the ground to the maximum height

→ The time to the maximum height is 1.50 seconds

→ The time to the ground = 2 × 1.50 = 3.0 seconds

The time to the ground is 3.0 seconds

C)

The maximum height of the ball is the vertical component of the displacement y

→ y = u sinФ t + g t²

The time for the maximum height is 1.50 seconds

→ g = -9.8 m/s² , u sinФ = 22 sin(42) , t = 1.5

- Substitute these values in the rule above

→ y = 22 sin(42)(1.5) - 4.9(1.5)²

→ y = 22.08 - 11.025

→ y = 11.06 m

The maximum height is 11.06 meters

D)

The horizontal displacement x is the product of the initial horizontal

component of the velocity and the time of the flight of the ball

→ x = (u cosФ) t

→ The initial horizontal component of velocity is 22 cos(42)

→ The time of the flight of the ball is 3 seconds

Substitute these values in the rule above

→ x = 22 cos(42) × 3

→ x = 49.05 m

The horizontal displacement of the kickoff is 49.05 meters

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.