Posted by **Charlie** on Saturday, March 29, 2008 at 3:20pm.

A cannon shoots a shell straight up. It reaches its maximum height, 1,051 feet, and splits into two pieces, one weighing 2 lb and the other 4 lb. The two pieces are observed to strike the ground simultaneously. The 4 lb piece hist the ground 1,608 feet away from the explosion (measured along the x axis). How long would it have taken the shell to return to the ground if it has not split? (answer to two decimal points)

- Physics -
**drwls**, Saturday, March 29, 2008 at 3:40pm
From the maximum height reached, you can determine the launch velocity

V = sqrt (2 g H) = 260 ft/s. Since the pieces at the ground arrive at the same time, they split in such a way that they acquire no vertical velocity components; otherwise one would go up and the other down. The time required for the pieces to fall after separating is T = sqrt (2*1051 ft/g) = 8.08 s

Since the the separation explosion occurred at the top of the trajectory and did not make the fall time different for the two pieces, the time to fall without the explosion would be the same as observed, 8.08 s. The info about how far away one piece landed is not needed.

- Physics -
**Charlie**, Saturday, March 29, 2008 at 4:47pm
OK thanks. There is another part of the question that asks for the momentum of the 4 lb piece and I can't figure out how to solve that.

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