# Calculating Initial Velocity for a Basketball Free Throw - Projectile Motion

• Devin Humphreys
In summary, to solve for the initial speed needed for a basketball player to make a three-pointer from a distance of 7.71 m, releasing the ball at a height of 1.93 m and shooting at a 45° angle, you can use the equations for horizontal and vertical motion to set up a system of equations with two unknowns (initial speed and time). By solving this system, you can find the initial speed needed for the player to make the basket.
Devin Humphreys

## Homework Statement

A basketball player practices shooting three-pointers from a distance of 7.71 m from the hoop, releasing the ball at a height of 1.93 m above ground. A standard basketball hoop's rim top is 3.05 m above the floor. The player shoots the ball at an angle of 45° with the horizontal. At what initial speed must he shoot to make the basket?

V0 = _______________ m/s

## Homework Equations

The four kinematic equations
Parabolic trajectory equation (simplified since x0 is taken to be 0 to

y = y0 + ((vy0 / vx0) * x) - ((g / (2vx02) * x2)

## The Attempt at a Solution

I know that the way the problem is set up it will take the same amount of time for the basketball to travel 7.71 meters horizontally as to reach the peak of its trajectory then go down to 3.05 meters in the y-direction for a displacement of 1.12 m. I don't know how to get from there to either finding the velocity directly or finding the time. If I can find the time, I can plug it and the y-displacement into the kinematic equation y = yi * t + (1/2)gt2 to get the initial velocity in the y direction, which I can apply basic trig to in order to find the total velocity. I just don't know how to find the time.

Devin Humphreys said:

## Homework Statement

A basketball player practices shooting three-pointers from a distance of 7.71 m from the hoop, releasing the ball at a height of 1.93 m above ground. A standard basketball hoop's rim top is 3.05 m above the floor. The player shoots the ball at an angle of 45° with the horizontal. At what initial speed must he shoot to make the basket?

V0 = _______________ m/s

## Homework Equations

The four kinematic equations
Parabolic trajectory equation (simplified since x0 is taken to be 0 to

y = y0 + ((vy0 / vx0) * x) - ((g / (2vx02) * x2)

## The Attempt at a Solution

I know that the way the problem is set up it will take the same amount of time for the basketball to travel 7.71 meters horizontally as to reach the peak of its trajectory then go down to 3.05 meters in the y-direction for a displacement of 1.12 m. I don't know how to get from there to either finding the velocity directly or finding the time. If I can find the time, I can plug it and the y-displacement into the kinematic equation y = yi * t + (1/2)gt2 to get the initial velocity in the y direction, which I can apply basic trig to in order to find the total velocity. I just don't know how to find the time.
You have two equations, one for horizontal motion and one for vertical. You have two unknowns, the initial angle and the time.
Two equations, two unknowns... solve.

haruspex said:
You have two equations, one for horizontal motion and one for vertical. You have two unknowns, the initial angle and the time.
Two equations, two unknowns... solve.

I have the initial angle (that's 45 degrees). What I don't have is the initial velocity, and that significantly complicates things. Is there some way I can use the initial angle to compensate even if both variables are found in both equations?

Devin Humphreys said:
I have the initial angle (that's 45 degrees). What I don't have is the initial velocity, and that significantly complicates things. Is there some way I can use the initial angle to compensate even if both variables are found in both equations?
Sorry, I meant initial speed, not angle. But it's the same deal. Let the time be t and the initial speed be v. What two equations can you write?

## 1. What is the formula for calculating initial velocity for a basketball free throw?

The formula for calculating initial velocity for a basketball free throw is V0 = d/[(sin 2θ)/g], where V0 is the initial velocity, d is the distance from the free throw line to the basket, θ is the angle of release, and g is the acceleration due to gravity.

## 2. How do you find the angle of release for a basketball free throw?

The angle of release can be found by using the formula θ = tan^-1[(Vy/Vx)], where Vy is the vertical component of velocity and Vx is the horizontal component of velocity.

## 3. What is the role of gravity in calculating initial velocity for a basketball free throw?

Gravity plays a crucial role in calculating initial velocity for a basketball free throw as it affects the vertical component of velocity. The acceleration due to gravity (g) is a constant value of 9.8 m/s^2 and is responsible for the downward motion of the basketball after it is released.

## 4. Can initial velocity be calculated for a basketball free throw without knowing the distance to the basket?

No, the distance to the basket is a crucial factor in calculating initial velocity for a basketball free throw. Without knowing the distance, the formula for initial velocity cannot be accurately calculated.

## 5. Is air resistance a factor in calculating initial velocity for a basketball free throw?

Air resistance can have a small impact on the initial velocity of a basketball, but it is often negligible. In most cases, it can be ignored when calculating initial velocity for a basketball free throw.

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