A basketballer is trying to make a shot in an international game from the 3 point line (6.75 m away from the hoop). The basketball hoop is 3.05 m above the floor, the basketballer’s hands are 1.95 m above the floor when they release the ball at an angle of 39° above the horizontal.
What velocity would the ball have to have been thrown with to reach and fall down into the hoop? And if the ball has a mass of 625 g what was the ball’s maximum potential energy during the throw?
To find the velocity needed for the ball to reach and fall into the hoop, we can use the projectile motion equations.
Step 1: Determine the horizontal and vertical components of the initial velocity.
The initial velocity can be broken down into its horizontal (Vx) and vertical (Vy) components. We can calculate these using the given angle and the total initial velocity.
Vx = V * cos(θ)
Vy = V * sin(θ)
Where:
V = initial velocity
θ = angle above the horizontal (39°)
Step 2: Find the time of flight.
The time it takes for the ball to reach the hoop is the time of flight (t). We can calculate this using the vertical component of the initial velocity.
t = (2 * Vy) / g
Where:
g = acceleration due to gravity (approximated as 9.8 m/s²)
Step 3: Calculate the horizontal distance traveled by the ball.
The horizontal distance traveled by the ball is given by multiplying the horizontal component of the initial velocity by the time of flight.
d = Vx * t
Step 4: Determine the initial velocity required for the ball to reach the hoop.
The initial velocity needed to reach the hoop is determined by satisfying the horizontal distance equation.
6.75 m = Vx * t
Step 5: Find the maximum potential energy of the ball during the throw.
The maximum potential energy is when the ball reaches its highest point. This occurs when the vertical velocity is zero. We can use the kinetic energy equation to find the potential energy.
Potential Energy = 0.5 * mass * (Vy)^2
Given:
Height of the hoop above the floor (h) = 3.05 m
Height of the basketballer's hands above the floor (H) = 1.95 m
Mass (m) = 625 g = 0.625 kg
Potential Energy = mass * g * (h + H)
Let's calculate the values.
Step 1:
Vx = V * cos(39°)
Vy = V * sin(39°)
Step 2:
t = (2 * Vy) / g
Step 3:
d = Vx * t
Step 4:
6.75 m = Vx * t
Step 5:
Potential Energy = mass * g * (h + H)
Solving these equations will give us the velocity required for the ball to reach the hoop and the maximum potential energy of the ball during the throw.
To find the velocity at which the ball would need to be thrown, we can use the principles of projectile motion. First, let's break down the given information:
- Distance from the 3-point line to the hoop = 6.75 m.
- Height of the hoop above the floor = 3.05 m.
- The basketballer's hand release height = 1.95 m.
- Angle of release above the horizontal = 39°.
To calculate the velocity, we need to determine the horizontal and vertical components separately. The initial velocity of the ball can be divided into two components: horizontal (Vx) and vertical (Vy).
The horizontal velocity (Vx) remains constant throughout the projectile motion since there is no horizontal acceleration. The formula for horizontal velocity is:
Vx = V * cos(θ)
where V is the initial velocity and θ is the angle of release.
We can find the horizontal velocity (Vx) by substituting the given values:
Vx = V * cos(39°)
Now, let's calculate the vertical velocity (Vy). The vertical velocity changes over time due to the acceleration due to gravity. The formula for vertical velocity is:
Vy = V * sin(θ)
Since we are interested in the velocity at which the ball reaches and falls into the hoop, we can use the following equation to find the time of flight:
-3.05 m = 1.95 m + (Vy * t) - (1/2 * g * t^2),
where g is the acceleration due to gravity (approximated as 9.8 m/s^2).
We can solve this equation for t (time of flight), which will allow us to find the vertical velocity component (Vy):
t = (Vy - sqrt(Vy^2 - 4 * (1/2 * g) * -1.1 m)) / (2 * (1/2 * g)),
where -1.1 m is the displacement between the release height and the hoop height.
Now, we can substitute the known values and rearrange the equation to solve for Vy:
t = (V * sin(39°) - sqrt((V * sin(39°))^2 - 2 * 9.8 m/s^2 * -1.1 m)) / (9.8 m/s^2)
Next, we can substitute this value of t in the equation for vertical velocity:
Vy = (3.05 m - 1.95 m + (1/2 * g * t^2)) / t
Now that we have both the horizontal and vertical velocity components (Vx and Vy), we can find the magnitude of the initial velocity (V) using the Pythagorean theorem:
V = sqrt(Vx^2 + Vy^2)
Finally, we can substitute the calculated values into the equation to determine the velocity at which the ball needs to be thrown.
To calculate the ball's maximum potential energy during the throw, we need to consider its height above the reference point (the floor). The potential energy formula is:
Potential energy = mass * gravity * height
Substituting the values given:
Potential energy = (0.625 kg) * (9.8 m/s^2) * (3.05 m - 1.95 m)
Simplifying this equation will give us the maximum potential energy.
Please plug in the given values and follow the aforementioned steps to calculate the velocity at which the ball needs to be thrown and the maximum potential energy of the ball.