A soccer player kicks a ball into the air at an angle of 36.0 Degrees above the horizontal. The initial velocity of the ball is +30.0 m/s.

1).How long is the soccer ball in the air
2).What is the horizontal distance traveled by the soccer ball?
3).What is the maximum height reached by the soccer ball?

1) Twice the time it takes for the vertical velocity component to become zero. 2*Vo*sin36/g = ___ s

2) (Vo^2/g)*sin72 = ___ m
3) (Vo*sin36)^2/(2g) = ___ m

To solve these problems, we can use the equations of motion for projectile motion. We'll break down each question step-by-step.

1) How long is the soccer ball in the air?
The motion of the soccer ball can be divided into two parts: the vertical motion and the horizontal motion. The time of flight of the soccer ball is the time it takes for the vertical motion to return to the ground.

The time of flight (T) can be calculated using the equation:
T = (2 * V * sinθ) / g
Where:
V is the initial velocity of the ball (30.0 m/s)
θ is the angle of projection (36.0 degrees)
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the values, we get:
T = (2 * 30.0 * sin(36.0)) / 9.8
T ≈ 3.07 seconds
Therefore, the soccer ball is in the air for approximately 3.07 seconds.

2) What is the horizontal distance traveled by the soccer ball?
The horizontal distance traveled by the soccer ball is given by the equation:
D = Vx * T
Where:
Vx is the horizontal component of the initial velocity (V * cosθ)
T is the time of flight calculated in the previous step.

Plugging in the values, we get:
D = (30.0 * cos(36.0)) * 3.07
D ≈ 76.89 meters
Therefore, the horizontal distance traveled by the soccer ball is approximately 76.89 meters.

3) What is the maximum height reached by the soccer ball?
The maximum height (H) reached by the soccer ball can be calculated using the equation:
H = (V^2 * sin^2θ) / (2 * g)

Plugging in the values, we get:
H = (30.0^2 * sin^2(36.0)) / (2 * 9.8)
H ≈ 21.48 meters
Therefore, the maximum height reached by the soccer ball is approximately 21.48 meters.

To solve this problem, we can use the basic principles of kinematics and projectile motion.

1) To find the time the soccer ball is in the air, we need to use the vertical component of the initial velocity. We can use the equation:
vf = vi + gt

where vf is the final vertical velocity (which will be zero when the ball reaches its maximum height), vi is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time. We can rewrite this equation to solve for time:

t = (vf - vi) / g

In this case, vi is 30.0 m/s and vf is 0 m/s. Plugging these values into the equation, we get:

t = (0 - 30.0) / 9.8

Calculating this, we find that the time the soccer ball is in the air is approximately 3.06 seconds.

2) To find the horizontal distance traveled by the soccer ball, we can use the equation:
d = vi * t * cos(θ)

where d is the horizontal distance, vi is the initial velocity, t is the time, and θ is the launch angle. In this case, vi is 30.0 m/s, t is 3.06 seconds, and θ is 36.0 degrees. We need to convert the angle to radians before using the equation:

θ_radians = θ * π / 180

d = vi * t * cos(θ_radians)

Plugging in the values, we get:

d = 30.0 * 3.06 * cos(36 * π / 180)

Calculating this, we find that the horizontal distance traveled is approximately 74.2 meters.

3) To find the maximum height reached by the soccer ball, we can use the equation:
h = vi^2 * sin^2(θ) / (2 * g)

where h is the maximum height, vi is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. In this case, vi is 30.0 m/s, θ is 36.0 degrees, and g is 9.8 m/s². We need to convert the angle to radians before using the equation:

θ_radians = θ * π / 180

h = vi^2 * sin^2(θ_radians) / (2 * g)

Plugging in the values, we get:

h = 30.0^2 * sin^2(36 * π / 180) / (2 * 9.8)

Calculating this, we find that the maximum height reached by the soccer ball is approximately 13.3 meters.

I NEED HELP TO FIND THE ANSWER YOU FOOLS