Proving Projectile Farthest Flight at 45° Angle

[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.

 

[dextercioby]

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

Daniel.

 

[HallsofIvy]

45 degrees assuming no air resistance!

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

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

position(t)= vcos(θ)t horizontally, vsin(θ)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 θ).

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

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

 

[LENIN]

Thenks! I got to the 2'nd step myself but the third one is really 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.