At the same moment from the top of a building 3.0 × 10² m tall, one rock is dropped and one is thrown downward with an initial velocity of 10 m/s. Both of them experience negligible air resistance. How much EARLIER does the thrown rock strike the ground? A) 0.95 s B) 0.86 s C) 0.67 s D) They land at exactly the same time.

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Quizplus · Accepted Answer

The dropped rock falls a distance of 3.0 × 10² m, while the thrown rock falls a distance of 3.0 × 10² m + 10 m/s × t, where t is the time it takes to reach the ground. Since both rocks experience negligible air resistance, they will both fall with the same acceleration due to gravity, g = 9.8 m/s². Therefore, the time it takes for the thrown rock to reach the ground can be found by solving the equation:3.0 × 10² m + 10 m/s × t = 3.0 × 10² m + 1/2 × g × t²Simplifying this equation, we get:1/2 × g × t² = 10 m/s × tDividing both sides by 1/2 × g, we get:t² = 20 m/s / gTaking the square root of both sides, we get:t = √(20 m/s / g) = 1.43 sTherefore, the thrown rock strikes the ground 1.43 s - 0.5 s = 0.93 s earlier than the dropped rock.

At the Same Moment from the Top of a Building