Fmonkey2001
New member
- Joined
- Aug 26, 2010
- Messages
- 1
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!
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!