Horizontal velocity of a basketball as it goes through the hoop

[yesmale4]

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

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

 

[PeroK]

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.

 

[PeroK]

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?

 

[yesmale4]

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

 

[yesmale4]

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

 

[PeroK]

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?

 

[yesmale4]

Why is the initial launch angle $45$ degrees?

Because that's the angle that is given me in the question

 

[PeroK]

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.

 

[yesmale4]

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?

 

[PeroK]

ohh i understand, do you have any idea how i can find it?

Do you know about energy? Or, only SUVAT formulas?

 

[yesmale4]

Do you know about energy? Or, only SUVAT formulas?

no we didnt learn energy yet

 

[PeroK]

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?

 

[yesmale4]

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

 

[kuruman]

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.

 

[PeroK]

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.