Posted by **leav** on Monday, December 5, 2011 at 3:14pm.

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
horizontalacceleration= F/m

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.

Interesting question.

- 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

thanks alot

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