Question

A shell is shot with an initial velocity of 12 m/s, at an angle of θ =53° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). 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 air drag is negligible?

A. Not enough information provided
B. The same distance as the vertically falling fragment
C. Depends on the mass of the fragments
D. Depends on the angle of explosion.

Answer 1

The other fragment will land at the same horizontal distance as the vertically falling fragment.

Step-by-step explanation:

To find the horizontal displacement of the shell when it explodes, we can use the concept of projectile motion. Since air drag is negligible and the fragments have equal mass, their horizontal displacements will be the same.

The horizontal displacement of a projectile can be calculated using the formula:

Δx = v*cos(θ)*t

where Δx is the horizontal displacement, v is the initial velocity of the shell, θ is the angle of projection, and t is the time of flight.

Since the fragments have equal mass, the other fragment will also land at the same horizontal distance from the gun as the vertically falling fragment. Therefore, the correct answer is B. The same distance as the vertically falling fragment.