A 40 kg artillery shell was fired from a cannon with an initial speed of 84 m/s

Question

A 40 kg artillery shell was fired from a cannon with an initial speed of 84 m/s at a 45° angle above the ground. Unfortunately, the shell was poorly made and exploded into two pieces 4 s after it was launched. One piece was five times heavier than the other. The heavier piece landed 600 m from the cannon. How far from the cannon (in meters) did the lighter piece land? HINT: the range of an artillery shell is given by the equation (v^2* sin(theata))/g , where v0 is the initial speed of the shell and ? is the launch angle.

Explanation / Answer

if the shell was correctly made its range would have been 84*84/10(check your formula it seems to be incorrect) =705.6 meters now it explodes into two with one being 5 times heavier than second or in pieces of 100/3 kg and 20/3 kg. now the center of mass will be at 705.6 m from cannon. so 40*705.6 = 600*100/3 + 20*R/3 [ r is the distance of smaller part of shell) R = 1233.6 m

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