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A projectile is launched from a point O on level ground at 45 degrees to the vertical with speed 5*sqrt(2) m/s. At the top of its path it breaks into two equal pieces. One piece drops straight down and lands 0.5 s after the break. The second piece lands t seconds after the break at a horizontal distance x from O. Take g = 10 m/s². Find t (in seconds).

1. 0.5
2. 1.0
3. 1.5
4. 2.0

Correct answer: 0.5

Solution

Both fragments leave the highest point at the same height with zero vertical velocity, so each takes the same time to fall; that time is 0.5 s.

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