To determine the release height of the basketball, we need to analyze the projectile motion of the ball.
First, we can break the initial velocity into its horizontal and vertical components. Since the ball is thrown horizontally, there is no initial vertical velocity. Therefore, the horizontal component of the velocity remains unchanged at 4.20 m/s throughout the motion, and the vertical component is 0 m/s.
Next, we can consider the vertical component of the motion. The ball experiences a constant acceleration due to gravity in the downward direction, equal to -9.8 m/s^2. We know that the displacement in the vertical direction can be calculated using the formula:
Δy = Vyi * t + (1/2) * a * t^2
where Δy is the vertical displacement, Vyi is the initial vertical velocity, t is the time of flight, and a is the acceleration due to gravity.
Since the initial vertical velocity is 0 m/s, the equation simplifies to:
Δy = (1/2) * a * t^2
Now we have to determine the time of flight, t. We can find it using the horizontal component of the motion and the known initial velocity in the horizontal direction.
The equation for the horizontal displacement, Δx, is:
Δx = Vxi * t
where Δx is the horizontal displacement and Vxi is the initial horizontal velocity.
In this case, the horizontal displacement is unknown, but we can calculate it using the given angle of 30.0 degrees.
The formula for the horizontal displacement is:
Δx = V * cos(θ) * t
where V is the initial velocity and θ is the angle with the horizontal.
Now, substituting the known values into the equation:
Δx = 4.20 m/s * cos(30.0°) * t
To find the time of flight, we can rearrange the equation:
t = Δx / (V * cos(θ))
Now that we have determined t, we can substitute it back into the equation for vertical displacement:
Δy = (1/2) * a * t^2
Plugging in the values:
Δy = (1/2) * (-9.8 m/s^2) * (Δx / (4.20 m/s * cos(30.0°)))^2
Simplifying the equation will give us the value of Δy, which represents the release height of the basketball.