Physics... Please help me

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? The fragment whose speed is zero, must have had momenum applied to it in the horizontal direction to "stop". That momenum is - 1/2 m vhorizontal, as it was going vhorizonal at the time. That means the other piece has 1/2 m vhorizontal applied to it, and it moves with m vhorizontal forward. So, since it's mass is 1/2 m, it now has twice the horizontal speed (2*20cos60). Figure the time it takes to get to the top. Then it takes the same time tofall, but the bulliet piecce is going twice as fast.


Using the equations of motion, the time to reach the top of the trajectory at a given angle (θ) is given by: t = (v_initial * sin θ) / g The time it takes to reach the ground is then 2t. The horizontal distance can then be calculated as: d = (2 * v_initial * cos θ * t) / g Therefore, in this case, the horizontal distance traveled is: d = (2 * 20m/s * cos 60 * (2 * (20m/s * sin 60) /9.81m/s^2) ) / 9.81m/s^2 = 171.42m

Answered by Joseph

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