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 recieve!

- 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

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