I am not sure if this is right, how can T = what that says... It would if the path was just one part, but it is two. You have to break it down into two parts don't you? Like the first part would be if you draw an imaginary line across the table, the whole motion from start to that line in the...
Ok I plug this into my clalc:
solve(.871+(6.425*sin(X))*(3.25/(6.425*cos(X)))-(9.8/2)*(.871/(6.425*cos(x)))^2=0,X)|X>0
It gives me the answer:
88.4233 degrees
I don't know if that is correct or what?
Ok, I got 13 degrees, but I think that is wrong. I am not sure how to do sec^2. You have to remember I am in america, my education isn't that great. I am useing a TI-89 to solve for theta in that equation, but like I said... I didn't even know there was a sec^2. Anyone got any other tips, thanks...
Well what I really need to know is the angle at which to place the gun so that it hits the target. Which means you would have to use the inverse of the trigometric functions (sin, cos, or tan). I think I might just make a program that finds the answer for me. THis isn't supposed to be an easy...
Can you give me the answer so I know that I am doing to right, the two angles? Because I have to show my work so even if you give me the answer it wouldn't matter. I just really need to get this right...
Velocity is not squared, that is a typo...
Also I know it deals with projectile motion, you can split it up into two separate problems, 1: from the gun to the level of the table, then 2: from that point to the floor. You would have to separate the projectile motion into X and Y motion and use...
http://chinkchinatown.com/extra.jpg
Ok as you can see the initial velocity is 6.425 m/s^2 and it is about .871m off the ground and about 3.25 m from the target. We have to find the angle at which to place the gun that shoots this ball to hit the target. This target it only like 1 inch at...