A golf ball was launched at 35.0m/s at an angle of 35.0 degrees above the horizontal

a) determine the horizontal and vertical components of the balls velocity
b)how long was the ball in the air
c) how far did the ball go
d) what was the maximum height

I don't know that's why I'm here

a) To determine the horizontal and vertical components of the ball's velocity, we need to use trigonometric functions.

The horizontal component of the velocity (Vx) can be found using the formula:

Vx = V * cos(θ)

Where V is the initial velocity of the ball and θ is the angle above the horizontal. Plugging in the given values:

Vx = 35.0 m/s * cos(35.0°)

Calculating this, we find:

Vx ≈ 35.0 m/s * 0.819 = 28.665 m/s

So, the horizontal component of the ball's velocity is approximately 28.665 m/s.

The vertical component of the velocity (Vy) can be found using the formula:

Vy = V * sin(θ)

Plugging in the given values:

Vy = 35.0 m/s * sin(35.0°)

Calculating this, we find:

Vy ≈ 35.0 m/s * 0.574 = 20.090 m/s

So, the vertical component of the ball's velocity is approximately 20.090 m/s.

b) To find how long the ball was in the air, we can use the vertical component of the velocity.

The time of flight can be determined using the formula for the vertical motion of an object:

Vy = Vy0 + g * t

Where Vy0 is the initial vertical velocity (20.090 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight. Since the ball starts at an angle of 35.0 degrees above the horizontal, the initial vertical velocity is positive.

Setting Vy to zero (as the ball reaches its peak and begins to descend), we have:

0 = 20.090 m/s - 9.8 m/s^2 * t

Solving for t, we get:

t ≈ 20.090 m/s / 9.8 m/s^2 ≈ 2.054 s

So, the time of flight for the ball is approximately 2.054 seconds.

c) To find how far the ball went horizontally, we need to use the horizontal component of the velocity and the time of flight.

The horizontal distance traveled (d) can be determined using the formula:

d = Vx * t

Where Vx is the horizontal component of the velocity (28.665 m/s) and t is the time of flight (2.054 s). Plugging in the values:

d ≈ 28.665 m/s * 2.054 s ≈ 58.832 m

So, the ball traveled approximately 58.832 meters horizontally.

d) To find the maximum height, we can use the vertical component of the velocity.

The maximum height (H) can be determined using the formula:

H = (Vy^2) / (2 * g)

Where Vy is the vertical component of the velocity (20.090 m/s) and g is the acceleration due to gravity (-9.8 m/s^2). Plugging in the values:

H ≈ (20.090 m/s)^2 / (2 * 9.8 m/s^2) ≈ 20.35 m

So, the maximum height reached by the ball is approximately 20.35 meters.