Solving an Initial Velocity Problem for a Basketball Player

[rockersdash]

1. Homework Statement

A basketball player is standing 9.5 m from the basket, which is at a height of 3.1m. she throws the ball from a initial height of 2m at an angle of 35 degrees above the horizontal. The ball goes through the basket. Determine intial velocity.





2. Homework Equations

Vx = Dx/t

v2^2 = v1^2 + 2ad

Dy = (V sine∅) - 0.5GT^2


3. The Attempt at a Solution

I took only half the parabola diagram and assumed v2 is 0 m/s

Vert

v1 = ?
v2 = 0 m/s
a = -9.8 m/s^2
max height = 1.1 m

I used the equation

v2^2 = v1^2 + 2ad

and got v1 as 4.64 m/s and from there on on I found initial velocity at an angle.

BUT the textbook gives a different answer (11m/s) as the initial velocity at an angle.

I can't figure out why I'm getting this wrong and my teacher sasys she will come back to me as she can't figure the mistake in mine... so help LOL

PS I assumed the max height of the parabola of the ball is 1.1 m otherwise I couldn't think of any other way.

Am I doing to this problem right? please post your work if any :)

 

[lewando]

You have already posted this problem and I have given you an approach to follow. Since you have posted slightly more information here, how about we stay with this one to keep this from getting out of control. Your original post:
https://www.physicsforums.com/showthread.php?t=636913

 

[lewando]

2. Homework Equations

Vx = Dx/t [this is good--what is V_x in terms of initial velocity, V?]

v2^2 = v1^2 + 2ad

Dy = (V sine∅) - 0.5GT^2 [this is also good--you are correctly using V_0,y = V sine∅. By the way, I now understand what you were trying to say by "max height": you meant to say "difference in height": (y - y0), which would be Dy in the above equation.]


3. The Attempt at a Solution

I took only half the parabola diagram and assumed v2 is 0 m/s [this is not good because you don't know how high the ball will go, nor at what point between the player and the basket this occurrs]

Vert

v1 = ?
v2 = 0 m/s
a = -9.8 m/s^2
max height = 1.1 m [again, this is not "max height" but the "difference in height"]

I used the equation

v2^2 = v1^2 + 2ad

and got v1 as 4.64 m/s and from there on on I found initial velocity at an angle.

BUT the textbook gives a different answer (11m/s) [which is correct]as the initial velocity at an angle.

I can't figure out why I'm getting this wrong and my teacher sasys she will come back to me as she can't figure the mistake in mine... so help LOL

PS I assumed the max height of the parabola of the ball is 1.1 m :uhh: otherwise I couldn't think of any other way.

Am I doing to this problem right? please post your work if any :)

Use:
Vx = Dx/t
Dy = (V sine∅) - 0.5GT^2

You have 2 equations and two unknowns (V and T). Should I just tell you that Vx = V cos∅, or was that already clear in your mind?