A shell is shot with an initial velocity 0 of 25 m/s, at an angle of 80 = 63° 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? Explosion

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**Projectile Motion and Explosion of a Shell**

A shell is launched with an initial velocity \(\vec{v}_0\) of 25 m/s at an angle of \(\theta_0 = 63^\circ\) relative to the horizontal. At the peak of its trajectory, the shell undergoes an explosion, splitting into two equal-mass fragments. One fragment comes to an immediate stop after the explosion and falls vertically. The task is to determine the horizontal distance from the launch point to the landing point of the other fragment, assuming a level surface and negligible air resistance.

**Illustration Explanation:**

The diagram below the explanation shows a projectile path curved upwards, representing the motion of the shell. The initial launch velocity \(\vec{v}_0\) is indicated by a pink arrow making an angle \(\theta_0\) with the horizontal. At the apex of the trajectory, the shell explodes, and the diagram illustrates this event with a small explosion icon.

- The fragment that stops immediately post-explosion is shown dropping directly downward.
- The other fragment continues along a new path away from the explosion point.

By using principles of conservation of momentum and the properties of projectile motion, one can calculate the distance the second fragment covers from the gun's position.
Transcribed Image Text:**Projectile Motion and Explosion of a Shell** A shell is launched with an initial velocity \(\vec{v}_0\) of 25 m/s at an angle of \(\theta_0 = 63^\circ\) relative to the horizontal. At the peak of its trajectory, the shell undergoes an explosion, splitting into two equal-mass fragments. One fragment comes to an immediate stop after the explosion and falls vertically. The task is to determine the horizontal distance from the launch point to the landing point of the other fragment, assuming a level surface and negligible air resistance. **Illustration Explanation:** The diagram below the explanation shows a projectile path curved upwards, representing the motion of the shell. The initial launch velocity \(\vec{v}_0\) is indicated by a pink arrow making an angle \(\theta_0\) with the horizontal. At the apex of the trajectory, the shell explodes, and the diagram illustrates this event with a small explosion icon. - The fragment that stops immediately post-explosion is shown dropping directly downward. - The other fragment continues along a new path away from the explosion point. By using principles of conservation of momentum and the properties of projectile motion, one can calculate the distance the second fragment covers from the gun's position.
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