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From the top of a 100 m tower, a particle is thrown vertically upward with speed 50 m/s at t = 0. Two seconds later, a second particle is projected vertically upward from the ground with the same speed 50 m/s. At what time after t = 0 are the two particles at the same height?

1. 11 s
2. 10 s
3. 5 s
4. 15 s

Correct answer: 11 s

Solution

Setting h1 = h2: 100 + 50t - 5t² = 50(t-2) - 5(t-2)². Expanding h2 = 70t - 120 - 5t². The t² and 50t terms cancel, leaving 100 = 20t - 120, so 20t = 220 and t = 11 s.

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