A shell is shot with initial velocity v0 of 20 m/s at an angle 60 degrees. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment has a speed of zero immediately after the explosion and falls vertically. How far from the gun does the other fragment land, assuming terrain is level and there is no air drag?
i don't know!
The Attempt at a Solution
i have no idea. i want to use v1f = (m1 – m2) / (m1+ m2) * v1i, but that leads to v1f = 0 which is clearly not correct in this situation -- why not? and if i use and v2f = 2m1 / (m1+ m2) * v1i, then i just get v2f = v1i, which is clearly not correct either. or maybe these are correct, like v1f is the fragment that falls vertically, and v2f just equals v1i somehow. but i still can't figure out how to get any distances out of this.
i really hate this problem, having looked at it for way too long, and i think my teacher will use something like this on the exam because he covered it again yesterday in class. this is what he wrote on the board:
vf = (m1+m2)/m2 * v0cos60
r - r0 = vft - 0.5gt^2
= Di - Hj
D = vft
-H = -0.5gt^2
D = vf sqrt(2H/g)
i'm sorry but this makes absolutely no sense to me. i vaguely recognize some of the projectile stuff here, but there's no way i could have done this on my own. in fact, i can't even do it with these notes in front of me. please help me. any advice on how to break this down. thanks.