A 1680- kg rocket is launched with a velocity v0 = 55 m/s against a strong wind. The wind exerts a constant horizontal force Fwind = 15600 N on the rocket. At what launch angle will the rocket achieve its maximum range?
physics - bobpursley, Monday, December 5, 2011 at 5:30pm
distance horzontal due to force= 1/2 F/m*t^2
distance horizontal= 55*cosTheta*t+1/2 F/Mass*t^2
Now, time in air:
hf=hi=0=55sinTheta*t-1/2 g t^2
or t= 110sinTheta/g
put that time in distance horizontal, then take the derivative of distance/dt and set to zero, solve for theta.
physics - leav, Friday, December 9, 2011 at 9:38pm
I DON'T UNDERSTAND.
WHAT IS: distance horzontal due to force= 1/2 F/m*t^2
and what do you mean by:
put that time in distance horizontal, then take the derivative of distance/dt and set to zero
if you can explain in more details it will help