A shell is fired from a gun with a muzzle velocity 466 m/s (Vo) at 57.7° above t

Question

A shell is fired from a gun with a muzzle velocity 466 m/s (Vo) at 57.7° above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming level terrain?

I have found the velocity of the "other" fragment (V1), V1 = 2Vxo = 2(466cos57.4)=502.134. I am stuck figuring out the distance of the shell from the gun to the point of explosion as well as the distance traveled by the gun after the explosion.

Explanation / Answer

time to reach maximum height = Vy/g = 466sin57.7 /9.8
= 40.19 sec
distance travelled by shell till explosion = Vx * time
= 466cos57.7 * 40.19
= 10007.63 m
applying conservation of momentum alonx x-direction (since no force acting along x -direction)
466cos57.7 = 0 + Vx since x-component of velocity is 0 for 1st shell
Vx = 249.00m/sec velocity of other shell along x-direction after explosion
now distance travelled by other shell = vx * time (taken by 1st shell to fall to ground)
= 249 * time
now we have to calculate the time to fall for 1st body
= (2 *Hmax/g )^1/2 [s = ut +1/2at^2]
now Hmax = (466sin57.7)^2 * (sin57.7)^2 / 2 *g
= 5655.62 m
hence time = 106.35 sec
hence distance travelled by other shell = 249 * 106.35 = 26482 m
total distnace = 26482 + 10007 =36489 m

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