A kickoff sends a football with an initial velocity of 25 m/s at an angle of 50 degrees above the horizontal.

a) find the x and y components of the velocity
b) find the time the ball is in the air
c) find the horizontal distance the ball travels before hitting the ground
d) find the maximum height of the ball

v⒳=v₀•cosα

v⒴=v₀•sinα
L=vₒ²•sin2α/g,
h= vₒ²•sin²α/2g,
t= 2vₒ•sinα/g

To find the x and y components of the velocity, we can use trigonometry.

a) The x-component of the velocity (Vx) can be found using the formula Vx = V * cos(θ), where V is the initial velocity (25 m/s) and θ is the angle above the horizontal (50 degrees).

Vx = 25 m/s * cos(50 degrees)
Vx = 25 m/s * 0.6428 ≈ 16.07 m/s

The y-component of the velocity (Vy) can be found using the formula Vy = V * sin(θ).

Vy = 25 m/s * sin(50 degrees)
Vy = 25 m/s * 0.766 ≈ 19.14 m/s

b) The time the ball is in the air can be found using the y-component of the velocity.

Using the formula: Vy = gt - Vy0, where g is the acceleration due to gravity (9.8 m/s^2), Vy0 is the initial vertical velocity (19.14 m/s), and t is the time.

0 = 9.8 m/s^2 * t - 19.14 m/s
19.14 m/s = 9.8 m/s^2 * t
t ≈ 1.95 seconds

c) The horizontal distance the ball travels before hitting the ground can be found using the horizontal component of the velocity and the time.

Using the formula: Distance = Velocity * Time.

Distance = 16.07 m/s * 1.95 seconds
Distance ≈ 31.31 meters

d) The maximum height of the ball can be found using the time it takes for the ball to reach its peak.

Using the formula: Vy = Vy0 + gt, where Vy0 is the initial vertical velocity (19.14 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the ball to reach its peak.

0 = 19.14 m/s + (9.8 m/s^2 * t_peak)
t_peak ≈ 1.95 seconds (from part b)

Using the formula: Maximum Height = Vy0 * t_peak - 0.5 * g * t_peak^2.

Maximum Height = 19.14 m/s * 1.95 seconds - 0.5 * 9.8 m/s^2 * (1.95 seconds)^2
Maximum Height ≈ 19.14 meters

Therefore,
a) The x and y components of the velocity are approximately 16.07 m/s and 19.14 m/s, respectively.
b) The time the ball is in the air is approximately 1.95 seconds.
c) The horizontal distance the ball travels before hitting the ground is approximately 31.31 meters.
d) The maximum height of the ball is approximately 19.14 meters.

a) To find the x and y components of the velocity, we can use trigonometry. The horizontal component (Vx) represents the velocity in the x-direction, and the vertical component (Vy) represents the velocity in the y-direction.

Vx = V * cos(θ)
Vx = 25 m/s * cos(50°)
Vx ≈ 25 m/s * 0.6428
Vx ≈ 16.07 m/s

Vy = V * sin(θ)
Vy = 25 m/s * sin(50°)
Vy ≈ 25 m/s * 0.7660
Vy ≈ 19.15 m/s

Therefore, the x component of the velocity (Vx) is approximately 16.07 m/s, and the y component of the velocity (Vy) is approximately 19.15 m/s.

b) The total time the ball is in the air can be determined by considering the vertical motion only. We can use the kinematic equation:

Vy = Voy - g * t

Where Vy is the final vertical velocity, Voy is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Since the ball is launched vertically, the final vertical velocity (Vy) will be -19.15 m/s (negative because it is moving downward). The initial vertical velocity (Voy) is 19.15 m/s. By substituting the values into the equation, we can solve for t.

-19.15 m/s = 19.15 m/s - 9.8 m/s^2 * t

Simplifying the equation, we get:

-9.8 m/s^2 * t = -38.3 m/s

Solving for t:

t = -38.3 m/s / (-9.8 m/s^2)
t ≈ 3.91 s

Therefore, the ball is in the air for approximately 3.91 seconds.

c) The horizontal distance the ball travels before hitting the ground can be found using the formula:

dx = Vx * t

Where dx is the horizontal distance, Vx is the horizontal component of velocity, and t is the time of flight.

Using the values we found earlier, we can substitute them into the formula:

dx = 16.07 m/s * 3.91 s
dx ≈ 62.84 m

Therefore, the ball travels approximately 62.84 meters horizontally before hitting the ground.

d) To find the maximum height of the ball, we can use the equation:

y = Voy * t + (1/2) * a * t^2

Where y is the height, Voy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.

At the maximum height, the vertical velocity is zero, so we can set Vy = 0. By substituting the known values, we can solve for t.

0 = 19.15 m/s - 9.8 m/s^2 * t

Simplifying the equation, we get:

9.8 m/s^2 * t = 19.15 m/s

Solving for t:

t = 19.15 m/s / 9.8 m/s^2
t ≈ 1.96 s

Now we can substitute the value of t into the equation for y to find the maximum height:

y = 19.15 m/s * 1.96 s + (1/2) * (-9.8 m/s^2) * (1.96 s)^2
y ≈ 19.15 m * 1.96 s - 9.8 m/s^2 * 1.96 s^2
y ≈ 37.58 m

Therefore, the maximum height of the ball is approximately 37.58 meters.