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## Homework Statement

A rocket is fired from from the ground at initial velocity of ##v_0## and at an angle ##\theta##. At its highest height it splits into 2 parts of equal masses. The first part is fired straight up and at velocity ##v_0/2##. Find the angle and intensity of the second part.

## Homework Equations

3. The Attempt at a Solution [/B]

At its maximum height the velocity has only the x component which is ##v_0cos\theta##. The momentum is ##p_0=mv_0cos\theta##. Since it has only the x component and the part which divides has the y component the second part must have an equal y component. Mening the ##v_{2y}=v_{1y}=v_0/2##. That would make the change in the y component of the momentum ##0##. Since that starting condition has total momentum in x direction the ##v_2## part must have the x component equal to the ##v_0cos\theta##. The intensity would then be ##\sqrt{v_{2y}^2+v_{2x}^2}##. Does this seem right couse when i use it in the problem given by the books where this intial velocity is ##1000## the angle ##60## i get ##v_y=500\sqrt{2}## and the book gives ##v_y=1118##?

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