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A student decides to conduct an experiment. He drops one ball from a balcony located 12.3 m above the ground and throws down a second ball at 6.35 m/s from a balcony located 25.0 m above the ground. Determine the velocity of each ball as it strikes the ground. a) 6.35 m b) 12.3 m c) 25.0 m d) 0.0 m

User JustSid
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1 Answer

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Final answer:

The velocity of ball A as it strikes the ground is approximately 15.47 m/s, while the velocity of ball B is approximately 16.94 m/s.

This correct answer is none of the above.

Step-by-step explanation:

To determine the velocity of the balls as they strike the ground, we can use the equations of motion.

For ball A, which is dropped from rest at a height of 12.3m, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (0 m/s for a dropped ball), a is the acceleration due to gravity (-9.8 m/s^2), and s is the distance traveled.

Plugging in the values, we get v^2 = 0 + 2(-9.8)(12.3), which simplifies to v^2 = -239.16. Since we only consider positive velocities, the velocity of ball A as it strikes the ground is approximately 15.47 m/s.

For ball B, which is thrown with an initial velocity of 6.35 m/s from a height of 25.0m, we can use the equation v^2 = u^2 + 2as. Plugging in the values, we get v^2 = (6.35)^2 + 2(-9.8)(25.0), which simplifies to v^2 = 286.75.

Therefore, the velocity of ball B as it strikes the ground is approximately 16.94 m/s.

This correct answer is none of the above.

User Milbrandt
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