# Finding the Initial Velocity of a Basketball Shot Using Freefall Equations

• VaioZ
In summary, the varsity player shoots a ball through the center of a ring at an angle of 50 degrees to the horizontal and an elevation of 2 meters. The ball leaves the player's hands at a speed of 12.446 meters per second.
VaioZ
Hello first time posting in the Physics forum!

1. Homework Statement

A varsity player is attempting to make a shot. The ball leaves the hands of the player at an angle of 50 degrees to the horizontal at an elevation of 2 meters above the floor. The skillful player makes the shot with the ball traveling precisely through the center of the ring (8 meters from the player and 3 meters above the floor). To loud cheers, calculate the speed at which the ball left the hands of the player. Sketch the problem!

Y=Vot-1/2g(t)^2
Voy = Vosintheta
Vox = Vocostheta

## The Attempt at a Solution

Vox = Vocostheta -> 8=Vocos(50degrees) -> Vo=12.446m/s
Then I plug it to X=Vot+1/2at^2 a is always 0 right? So it becomes X=Vot -> X=12.446(t) -> t=(0.643s)
Then I plug the time to the y equation of free fall.. Delta Y is 1 right? Because 3-2=1
Y=Vot-1/2gt -> 1=Vo(0.643)-1/2(9.8)(0.643)^2 Vo=4.7059m/s
y
I'm not sure if my answer is right.. if not can someone guide me...
If it is wrong sorry If I got a wrong answer.. This is why I post in this forums.

Hello VZ, welcome to PF !

Always a good idea to include dimensions in your calculations -- and to check them. In your case, you go off the rails in the very first line (sorry to say):

## 8 \;{\rm m} = V_{0,x} = V_0 \cos\theta \;{\rm m/s}## can never be right !

Thing to do is write expressions for x(t) = 8 m and y(t) = 3 m, and eliminate t. Best to work with symbols until you have an expression for ##\theta##. Then check the dimensions and do the calculator work.

BvU said:
Hello VZ, welcome to PF !

Always a good idea to include dimensions in your calculations -- and to check them. In your case, you go off the rails in the very first line (sorry to say):

## 8 \;{\rm m} = V_{0,x} = V_0 \cos\theta \;{\rm m/s}## can never be right !

Thing to do is write expressions for x(t) = 8 m and y(t) = 3 m, and eliminate t. Best to work with symbols until you have an expression for ##\theta##. Then check the dimensions and do the calculator work.

I don't know what you mean by dimension buuuuuuut..
ΔX = Vo(t)
ΔY = Vo(t)-1/2(g)(t)^2

Do you mean that I should experiment with this two formulas? To get Vo? or to get t?

VaioZ said:
Vox = Vocostheta -> 8=Vocos(50degrees) -> Vo=12.446m/s
Then I plug it to X=Vot+1/2at^2 a is always 0 right? So it becomes X=Vot -> X=12.446(t) -> t=(0.643s)
Then I plug the time to the y equation of free fall.. Delta Y is 1 right? Because 3-2=1
Y=Vot-1/2gt -> 1=Vo(0.643)-1/2(9.8)(0.643)^2 Vo=4.7059m/s
y
I'm not sure if my answer is right.. if not can someone guide me...
If it is wrong sorry If I got a wrong answer.. This is why I post in this forums.
First - you have equated horizontal velocity to horizontal displacement . This is wrong .
So your second and third equations automatically become wrong .

Qwertywerty said:
First - you have equated horizontal velocity to horizontal displacement . This is wrong .
So your second and third equations automatically become wrong .

Yeap I realized that after sir BvU post. Sooooo if it is okay can you guide me in this question?

VaioZ said:
Yeap I realized that after sir BvU post. Sooooo if it is okay can you guide me in this question?
Consider two unknowns v and t . Obviously v represents velocity of object , and t the time taken to reach the hoop .

What you'll need to do is form two equations . Use the fact that horizontal velocity of object is v*cos(θ) and vertical v*sin(θ) .

Can you manage the two equations ?

Qwertywerty said:
Consider two unknowns v and t . Obviously v represents velocity of object , and t the time taken to reach the hoop .

What you'll need to do is form two equations . Use the fact that horizontal velocity of object is v*cos(θ) and vertical v*sin(θ) .

Can you manage the two equations ?

Okay will do! I'll update you if I'm done. Thank you

Qwertywerty said:
Consider two unknowns v and t . Obviously v represents velocity of object , and t the time taken to reach the hoop .

What you'll need to do is form two equations . Use the fact that horizontal velocity of object is v*cos(θ) and vertical v*sin(θ) .

Can you manage the two equations ?

SIR I THINK I GOT IT. Okay here goes

ΔX = Vox(t) right? bc a is always 0
ΔX/Vox = t but we know that Vox=VoCos50 and ΔX = 8
so 8/Vocos50 = t
Voy=Vosin50 right?
So
Δy=Voy(t)-1/2g(t)^2
3-2=(Vosin50)(Vocos50)-1/2(9.8)(Vocos50)^2

VaioZ said:
Δy=Voy(t)-1/2g(t)^2
3-2=(Vosin50)(Vocos50)-1/2(9.8)(Vocos50)^2
You need to substitute correctly for t .

Qwertywerty said:
You need to substitute correctly for t .

typo 8/Vocos50 but it is right the context is still there..

Yes it is .

VaioZ said:
I don't know what you mean by dimension buuuuuuut..
ΔX = Vo(t)
ΔY = Vo(t)-1/2(g)(t)^2

Do you mean that I should experiment with this two formulas? To get Vo? or to get t?
No experimenting: you want to manipulate these two equations with two unknowns in such a way that you get one equation with one unknown, namely ##\theta##

The way you write them is confusing -- one Vo with two different meanings, and Vo(t) makes it look as if Vo is a function of t --$$\Delta X = v_0 \cos\theta\; t \\ \Delta Y = v_0 \sin\theta\; t - {1\over 2} gt^2$$ is a lot clearer. See also here

 oops, I'm lagging. This is in response to post #3 -- a bit late.

Qwertywerty

## What is the "Freefall basketball problem"?

The "Freefall basketball problem" is a physics problem that involves calculating the height and time of a basketball in freefall. It is often used as an example to demonstrate the principles of freefall and projectile motion.

## What are the key components of the "Freefall basketball problem"?

The key components of the "Freefall basketball problem" are the initial height of the basketball, the acceleration due to gravity, and the initial velocity of the basketball. These values are used to calculate the height and time of the basketball in freefall.

## How do you solve the "Freefall basketball problem"?

To solve the "Freefall basketball problem", you first need to gather the necessary information, including the initial height, acceleration due to gravity, and initial velocity. Next, you can use the formula h = h0 + v0t + (1/2)at^2 to calculate the height of the basketball at any given time. You can also use the formula v = v0 + at to calculate the velocity of the basketball at any given time.

## What are some real-life applications of the "Freefall basketball problem"?

The "Freefall basketball problem" has many real-life applications, such as predicting the trajectory of a basketball shot, calculating the time it takes for a ball to drop from a certain height, and understanding the physics of freefall and projectile motion.

One common misconception about the "Freefall basketball problem" is that the basketball will continue to accelerate until it reaches the ground. In reality, the basketball will reach a maximum height and then start to decelerate due to air resistance. Another misconception is that the basketball will fall straight down, when in fact it will follow a curved path due to the force of gravity.

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