- #1
Coco12
- 272
- 0
Homework Statement
A cannon fires a cannonball 500 m down range when set at a 45 degree angle.At what velocity does the cannonball leave the cannon?
Homework Equations
Vyi=sin45 * Vi
Vxi=cos45* Vi
The Attempt at a Solution
I don't know how to do this question?? Can someone help me?
If u had time u could solve it but how to find that when u don't even know Vyi?
I found this solution online.. I understand everything until line number 10. How did they get cos 45? Any help would be greatly appreciated
Vyf = Vi sin 45+ at,
at = - Vi sin45
-gt = - Vi sin 45
t = Vi sin 45/g
And we defined t = 1/2 T or T = 2t
T = 2*Vi sin 45/g
Lets pull back up [eq1] and plug in this last equation into it.
Vi cos 45 = R / T
Vi cos 45 = R / (2*Vi sin 45/g)
Vi^2 = g*R / (2*sin 45 *cos 45)
Vi = sqrt (g*R / (2*sin 45 *cos 45))
There you go, now just plug in your numbers.
Vi = sqrt (9.81 m/s^2 *500 m / (2*sin 45 *cos 45))
Vi = 70 m/s
Last edited: