Posted by Jerry on Thursday, August 26, 2010 at 3:58pm.
A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45' and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the intitial speed of the ball, and how high does it rise?
So far i have got a few formulas, but i am still at a loss at where to start and when i start to put my formulas together i start to get zeros because everything is canceling out.
Y= h + (V[sub]0[/sub]Sin(x))t  (1/2)gt^2)
T= (V[sub]0[/sub]Sin(x))/g
X=V[sub]0[/sub]Cos(x)
Y=V[sub]0[/sub]Sin(x)
Tan(x)=(Sin(x))/(Cos(x))
Then i have the formula for posistion vector, velocity vector, and acceleration vector.
Where; I, J and K are the vectors
r(t)=(V[sub]0[/sub]Cos(x))tI + (h + (V[sub]0[/sub]Sin(x))t  (1/2)gt^2)
v(t)= (V[sub]0[/sub]Cos(x))I + ((V[sub]0[/sub]Sin(x))  gtJ)
a(t)= g J
Thanks for any help i receive!

Calculus  drwls, Thursday, August 26, 2010 at 6:39pm
For a ball caught at the same altitude that it is hit, the horizontal distance travelled is
X = (Vo^2/g)sin(2A)
where A is the "launch" angle from horizontal.
In your case 2A = 45 degrees, so
Vo^2 = g X
Solve for Vo.
The height it rises is
H = (Vo sinA)^2/g = Vo^2/(2g) = X/2