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2

If your textbook actually derives $(2)$ as the motion of a thrown object, throw it away. The general trajectory of an object thrown from $(0,0)$ at angle $\theta$ is $$ y(x) = x\tan(\theta) - \frac{gx^2}{2v^2}(1+\tan(\theta)^2)\tag{1}$$ and now you say you "impose $\Delta = 0$". Let's analyse that "imposing" a bit more carefully, it is equivalent to $$ v^4 ...


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It would still be here, it would just be turned inside out.


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You can approach this in a couple of different ways. 1) The motion of the center of mass of the particles will follow the same path the grenade would have followed if it had not exploded. 2) The total momentum of the two pieces immediately after the explosion will be equal to the momentum of the grenade immediately before the explosion. Enjoy the algebra ...


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@ Nick Basically you are confusing two different coordinate systems. In the coordinate system where you decomposed g into two components, there the question of x and y doesn't seem valid because you have transformed the cartesian coordinate system into a different one. You would have new x and new y there and you have to stick with one while solving the ...


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For the second part to travel down the path of the full grenade, it must have a velocity $-V_x$. By taking momentum into account, it can then be shown that the other part must have a horizontal velocity $3V_x$. Momentum is conserved, even though kinetic energy is not. You can then use these new velocities to calculate the distance between the two points ...



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