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.
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 dont know why I am now getting different valudes for t.
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.
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 =...
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...
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...