Calculus- Projectile Motion

[Fmonkey2001]

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:1tv0dspr]0[/sub:1tv0dspr]Sin(x))t - (1/2)gt^2)
T= (V[sub:1tv0dspr]0[/sub:1tv0dspr]Sin(x))/g
X=V[sub:1tv0dspr]0[/sub:1tv0dspr]Cos(x)
Y=V[sub:1tv0dspr]0[/sub:1tv0dspr]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:1tv0dspr]0[/sub:1tv0dspr]Cos(x))tI + (h + (V[sub:1tv0dspr]0[/sub:1tv0dspr]Sin(x))t - (1/2)gt^2)
v(t)= (V[sub:1tv0dspr]0[/sub:1tv0dspr]Cos(x))I + ((V[sub:1tv0dspr]0[/sub:1tv0dspr]Sin(x)) - gtJ)
a(t)= -g J

Thanks for any help i recieve!

 

[BigGlenntheHeavy]

$\displaystyle The \ path \ of \ the \ ball \ is \ given \ by:$

$\displaystyle r(t) \ = \ [v_0cos(45^0)]ti+[3+[v_0sin(45^0)t]-16t^2]j, \ h \ = \ 3.$

$\displaystyle = \ \bigg(\frac{tv_0}{\sqrt2}\bigg)i+\bigg(3+\frac{tv_0}{\sqrt2}-16t^2\bigg)j$

$\displaystyle Now, \ x(t) \ = \ 300 \ when \ y(t) \ = \ 3, \ ergo;$

$\displaystyle \frac{tv_0}{\sqrt2} \ = \ 300 \ and \ 3+\frac{tv_0}{\sqrt2}-16t^2 \ = \ 3.$

$\displaystyle Hence, \ t \ = \ \frac{300\sqrt2}{v_0} \ \implies \ \frac{300\sqrt2}{v_0}\bigg(\frac{v_0}{\sqrt2}\bigg)-16\bigg(\frac{300\sqrt2}{v_0}\bigg)^2 \ = \ 0$

$\displaystyle 300 \ = \ \frac{16(300)^2(2)}{v_0^2}, \ v_0^2 \ = \ 32(300), \ v_0 \ = \ \sqrt{9600} \ = \ 40\sqrt6 \ \dot= \ 97.98 \ ft/s$

$\displaystyle Now, \ can \ you \ finish \ up?$