Projectile Motion Problems: Calculating Maximum Angles and Heights

[hancyu]

r(t)^2 = vcos(t)^2 + [vsin(t) - 1/2(gt^2)]^2

then what?

 

[Dick]

You really can't just keep asking me questions without putting any thought in before hand, ok? I already told you. You have to make sure that the derivative of r(t)^2 is never zero. Then it can't come back. And I wouldn't use t for both time and angle, it's going to be confusing. Differentiate it, ok? With respect to t. Not t. Told you it would be confusing.

 

[lorenzom21]

No, it's not 45 degrees. If the launch angle is theta then you work out x(t) and y(t). The distance squared from the launch point (0,0) is r(t)^2=x(t)^2+y(t)^2. You want to take the derivative of r(t)^2 with respect to t and then figure out for which angles theta that expression has a zero derivative. These are the ones where it comes back towards you.

Dick, why is it that I have to find the derivative of r(t)^2? the distance is r(t) so shouldn't I derive r(t) with respect to t and not r(t)^2?

 

[Dick]

Dick, why is it that I have to find the derivative of r(t)^2? the distance is r(t) so shouldn't I derive r(t) with respect to t and not r(t)^2?

If r(t) is increasing/decreasing then r(t)^2 is increasing/decreasing. You can do either one. I just suggest r(t)^2 because it's easier. It doesn't have a square root in it.