# What's the proof

1. Mar 8, 2005

### LENIN

I know that a projectile flyes the farthest, when it is louncht under a 45 degree angel. But I don't know how to proofe this. Any help would be apreeciated.

2. Mar 8, 2005

### dextercioby

Well,then can u write down the equations of motion...?

Daniel.

3. Mar 8, 2005

### HallsofIvy

Staff Emeritus
45 degrees assuming no air resistance!

If the angle is &theta; degrees, initial speed v, then the horizontal and vertical components of initial velocity are vcos(&theta;) and vsin(&theta) respectively.

acceleration is 0 horizontally, - g vertically so
velocity (t)= vcos(&theta;) horizontally, vsin(&theta;)- gt vertically.

position(t)= vcos(&theta;)t horizontally, vsin(&theta;)t- (1/2)gt2 vertically. (taking (0,0) as starting point.)

1. Solve for the time of impact (i.e. set height= 0, find non-zero t solution- it will depend upon &theta;).

2. Find horizontal distance at that time (again, it will depend upon &theta;).

3. Find the value of &theta; that maximizes that (differentiate with respect to &theta;, set equal to 0).

4. Mar 8, 2005

### LENIN

Thenks! I got to the 2'nd step myself but the third one is realy helpful. I didn't think about useing differentilas, becouse I only started takeing them a few weaks ago and I still don't find my way around very well. Thenks agein.