A shell is fired from a gun with a muzzle velocity of 37.2 m/s, at an angle of q = 60.4° with the horizontal.

Question

At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-38). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible?

Answer 1

First, determine the x and y components of the velocity at the time the gun is fired:

vx=v*cos60.4=18.4 m/s
vy=v*sin60.4=32.3 m/s

Next, calculate the time it takes for the y component of the shell to reach 0 (the top of the trajectory):

delta(vy)= ay*t where ay=the acceleration in the y direction (gravity)
or, t=delta(vy)/ay = 32.3/9.8 = 3.3 s

That means the shell will travel a total of 6.6 s before gravity pulls it back to the ground.

Now, for the first half of the trajectory, the shell moves horizontally a distance of 18.4 m/s * 3.3 s = 60.7 m

However, since the shell splits equally (loses half of its mass), the conservation of momentum tells us that the horizontal velocity of the shell has to double (since there are no momentum losses, as the fragment of the shell that falls off has no forward velocity).

So, for the second half of the trajectory, the shell travels twice as far, or 121.4 m.

So, the total distance is 60.7+121.4 = 182.1 m

Hope this helps