Horizontal velocity of a basketball as it goes through the hoop

In summary: The second step is to use the kinematic equation to solve for the ##v_x## and ##v_y##. Again, I'll give you a hint:$$v_x = -v_y$$
  • #1
yesmale4
41
1
Homework Statement
A basketball player who is 2.00 m tall is standing 8 m from the basket and throws a ball. The ball enters the basket (without striking the backboard) at θ=45° with the horizontal.
Assume that the basket height is 3.05 m.
Relevant Equations
4 constant acceleration equations
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hey i don't understand why my answers are incorrect, here is my solution i would like if someone can help me understand what I am doing worng and how i should solve this problem
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  • #2
First, I'd like to understand your solution. I can see numbers, some of which are legible, but no explanation of what you're doing.
 
  • #3
Here's my attempt at a solution:
##6.31725 = 1.245631t + 32.222v_0##
##v_1 = 55.125 - 18.372 = 36.76753##

Now, can you tell me what am I doing wrong?
 
  • #4
PeroK said:
First, I'd like to understand your solution. I can see numbers, some of which are legible, but no explanation of what you're doing.
yes of course, first of all i split the motion to two axes - X and Y
X:
v0 = v0cos45
V = v0cos45
t = ?
a=0
x =8m

Y:
v0= v0sin45
V = ?
a = -9.82m/s
t = ?
y=3.05m

t=(x-x0 )/v0
after i do this i use y=y0+v0*t+1/2*g*t^2
and than i found v0=9.509
 
  • #5
PeroK said:
Here's my attempt at a solution:
##6.31725 = 1.245631t + 32.222v_0##
##v_1 = 55.125 - 18.372 = 36.76753##

Now, can you tell me what am I doing wrong?
no I am sorry i don't understand what you are doing worng
 
  • #6
yesmale4 said:
yes of course, first of all i split the motion to two axes - X and Y
X:
v0 = v0cos45
V = v0cos45
Why is the initial launch angle ##45## degrees?
 
  • #7
PeroK said:
Why is the initial launch angle ##45## degrees?
Because that's the angle that is given me in the question
 
  • #8
yesmale4 said:
Because that's the angle that is given me in the question
That's the angle at which the ball goes into the hoop; not the angle with which it's launched.
 
  • #9
PeroK said:
That's the angle at which the ball goes into the hoop; not the angle with which it's launched.
ohh i understand, do you have any idea how i can find it?
 
  • #10
yesmale4 said:
ohh i understand, do you have any idea how i can find it?
Do you know about energy? Or, only SUVAT formulas?
 
  • #11
PeroK said:
Do you know about energy? Or, only SUVAT formulas?
no we didnt learn energy yet
 
  • #12
yesmale4 said:
no we didnt learn energy yet
You have three unknowns: initial velocity, initial angle and time. So, I guess, you need three equations:

The x-displacement and y-displacement are two equations.

The third equation is that ##v_x = -v_y## when the ball enters the hoop. That's using the ##45## degrees.

Can you make progress from that?
 
  • #13
PeroK said:
You have three unknowns: initial velocity, initial angle and time. So, I guess, you need three equations:

The x-displacement and y-displacement are two equations.

The third equation is that ##v_x = -v_y## when the ball enters the hoop. That's using the ##45## degrees.

Can you make progress from that?
but X we have its equal to 8 and Y is equal to 1.05 , about the degrees i still don't understand how to find it
 
  • #14
You know that the time of flight is the distance traveled in each direction divided by the average velocity in that direction $$t_{\!f}=\frac{\Delta x}{\bar v_x}=\frac{\Delta y}{\bar v_y}$$ You also know the kinematic equation $$2(-g)\Delta y=v_{\!fy}^2-v_{0y}^2$$ All you need to do is
1. Find expressions for the average velocities in terms of their final and initial values.
2. Put it together. Substituting @PeroK's hint (his third equation), gives you a system of two equations and two unknowns, the initial components of the velocity.
 
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  • #15
yesmale4 said:
but X we have its equal to 8 and Y is equal to 1.05 , about the degrees i still don't understand how to find it
The first step, I suggest, is write the equations for the ##x## and ##y## displaments. The ##x## displacement is the simplest, so I'll give you that as a further help:
$$x = v_0 t \cos \theta$$Where ##v_0## is the initial velocity and ##\theta## is the launch angle.
 

FAQ: Horizontal velocity of a basketball as it goes through the hoop

What is the horizontal velocity of a basketball as it goes through the hoop?

The horizontal velocity of a basketball as it goes through the hoop depends on various factors such as the initial velocity, angle of release, and air resistance. However, on average, the horizontal velocity can range from 10-15 meters per second.

How does the horizontal velocity affect the trajectory of the basketball?

The horizontal velocity of a basketball is responsible for the horizontal distance it travels as it goes through the hoop. A higher horizontal velocity will result in a longer horizontal distance covered by the basketball before it reaches the hoop.

Is the horizontal velocity the same throughout the entire flight of the basketball?

No, the horizontal velocity of a basketball changes throughout its flight due to factors such as air resistance and the gravitational pull of the Earth. It is highest at the initial release and decreases as the basketball moves towards the hoop.

How does the horizontal velocity of a basketball affect its chances of going through the hoop?

The horizontal velocity of a basketball does not directly affect its chances of going through the hoop. However, a higher horizontal velocity can result in the basketball covering a longer distance, giving it a better chance of going through the hoop.

Can the horizontal velocity of a basketball be controlled by the player?

Yes, the horizontal velocity of a basketball can be controlled by the player through their technique and strength. A player can increase the horizontal velocity by using more force and releasing the ball at a higher angle, or decrease it by using less force and releasing the ball at a lower angle.

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