A projectile fired into the air, explodes, and splits into two halves of equal mass that hit the ground at the same time. If the projectile had not exploded, it would have landed at point X, which is a distance R to the right of the launch point. After the explosion, one of the halves lands at point Y, which is a distance 2R to the right of the launch point. If air resistance is negligible, where did the other half land

The other half landed at the launch point. We were told that after the

explosion of the projectile there was a split occurred in which one of the

halves landed at the point Y, which is a distance of 2R to the right side of the

launch point.

This means that the only way one half of the object would have gone double

would be definitely if the original distance of R has a velocity which was two

times that of the initial velocity.

We can conclude that the other half fell straight down and landed at the

initial velocity point which is at launch point.

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At the Launch point

Explanation:

We are told that after the explosion, one of the halves landed at the point Y, which is a distance of 2R to the right side of the launch point.

This means that the only way one half of the object would have gone double the original distance of R is if it's velocity was two times that of the initial velocity. Furthermore, the only way that will make sense is if the other half fell straight down. So In conclusion, if air resistance is neglected, the other half will land at the initial velocity point which is at launch point.