A shell is shot with an inital velocity of 20m/s at an angle 60 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment whose speed is immediately after the explosion zero falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and the air drag is negligible?
Would I use v^2 =v_o^2 +2ad? and solve for d? but would I also use F=ma to find a?
1 answer
First break up V initial into components. Then you must calculate the time it takes for the shell to reach it's peak using Vy=V0y-gt (should get 1.77s). Then, find displacement in the x before the explosion happens using x=Vox*t (since there's no acceleration in the x-direction, the product to the right of the sum goes to zero). We get 17.7m in the x. Then we can use the formula for center of mass to find where the other fragment lands.
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