Need work checked(dyanmics projectile problem)

[LightMech]

First time poster, site looks very informative so I thought I would make a post. This is for Mechanics class and now covering Dynamics. This is my take home problem.

A basket ball player shoots when she is 5 m from the backboard. Knowing that the ball has an intial velocity Vo at an angle of 30⁰ with the horizontal, determine the value of Vo when d is equal to (a) 228 mm, (b) 420 mm.

Homework Statement

given info:
x: player is 5M from back board
y: initial height of ball 2m from floor, hoop is 3.048m from floor
angle: 30⁰
d: this is the distance from the back board or offset

Homework Equations

x = Vo cos(30)*t
y = Yo + Vo*sin(30)t - 1/2gt^2

The Attempt at a Solution

x = 5m - .288m = 4.772m
4.772m = Vo*cos(30)*t
Vo = 4.772/cos(30)

y = 2m + (4.772m/cos(30))*sin(30)t - 1/2(9.81)t^2
t = .987s

Vo = 4.772/cos(30)*(.987s)
= 5.44m/s

Using same method
b = 5.13m/s
Found a picture for the problem on the net for better clarity:
http://books.google.com/books?id=A_...esnum=3&ved=0CBEQ6AEwAg#v=onepage&q=&f=false"

 

[LightMech]

So its been several days now and no response. Perhaps I would have had better luck under the Physics section? If I did not explain the problem clearly I posted a link to the problem which includes a picture. I am pretty confident my work is correct but would feel a lot better if someone else can confirm this work accurate.

 

[PhanthomJay]

Equations look good, but check your math and you have a typo , should be Vo = 4.772/cos(30)*t (you left out the 't'). What value are you using for y?

 

[LightMech]

I am solving for 't', so y = 0. But I use 2m for my Y0.
Here is a little more detail in my steps:
x = Vo cos(30)*t
4.772m = Vo*cos(30)*t
Vo = 4.772/cos(30)t
y = Yo + Vo*sin(30)t - 1/2gt^2
0 = 2m + (4.772m/cos(30))*sin(30)t - 1/2(9.81)t^2
(4.9)t^2 = (4.775)t^1/2
t = .987s
Vo = 4.772/cos(30)*(.987s)
= 5.44m/s

 

[LightMech]

I am not sure about my value for 't'. I take the square root of both sides and solve for t, I get .987 when 4.9 over 4.775 or 1.013 when 4.775 over 4.9.

 

[PhanthomJay]

I am solving for 't', so y = 0.

y is not 0 when the ball hits the basket.

But I use 2m for my Y0.
Here is a little more detail in my steps:
x = Vo cos(30)*t
4.772m = Vo*cos(30)*t
Vo = 4.772/cos(30)t
y = Yo + Vo*sin(30)t - 1/2gt^2
0 use correct value for y[/color]= 2m + (4.772m/cos(30)t[/color])*sin(30)t - 1/2(9.81)t^2

you keep making the same error by omitting the t [/color]I've added in red.

 

[LightMech]

Geeze I can't believe I mist that. Okay so:
Yo = 2m
Yf = 3.048m
3.048m = 2m + (4.772m/cos(30)t)*sin(30)t - 1/2(9.81)t^2
t = 1.7s
Vo = 4.772/cos(30)*(1.7s)
V0 = 9.37m/s

 

[PhanthomJay]

Geeze I can't believe I mist that. Okay so:
Yo = 2m
Yf = 3.048m
3.048m = 2m + (4.772m/cos(30)t)*sin(30)t - 1/2(9.81)t^2
t = 1.7s
Vo = 4.772/cos(30)*(1.7s)
V0 = 9.37m/s

you seem to be having trouble with your math

1.048 = (4.772/cos30)sin30 -1/2(9.81)t^2

solve for t = ?

 

[LightMech]

Ok, this is what I am doing:
4.9t^2= -1.048+(4.772/cos30)sin30 t's cancel on right side?
taking square toot to both sides and solving for t:
2.21t= 1.306
t=.59s

I am moving t^2 to the left side, I don't know why I am now getting different valudes for t.

 

[PhanthomJay]

Ok, this is what I am doing:
4.9t^2= -1.048+(4.772/cos30)sin30 t's cancel on right side?
taking square toot to both sides and solving for t:
2.21t= 1.306
t=.59s

I am moving t^2 to the left side, I don't know why I am now getting different valudes for t.

Yes, that looks better. I'm not sure what you were doing before.

 

[LightMech]

Thank you Phantom, I wasnt taking care of my 't' because I had it written incorrectly on a different set of notes. If I had took the time to think about it, I probably would have corrected it.