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A ball with a mass of 2.0 kg is dropped from the top of a building and hits the ground with a velocity of -50.0 m/s. a) What is height of the building? b) If the building were twice as tall, calculate how this would change the final velocity of the ball.​

User Ymoreau
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Final answer:

The height of the building is 127.6 meters. Doubling the height of the building would not change the final velocity of the ball.

Step-by-step explanation:

a) What is the height of the building?

To calculate the height of the building, we can use the equation for the final velocity of an object dropped from rest:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration due to gravity (-9.8 m/s^2), and s is the height of the building. Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Substituting the given values, we have:

s = (-50.0^2 - 0^2) / (2 * -9.8) = 127.6 meters

Therefore, the height of the building is 127.6 meters.

b) How would doubling the height of the building change the final velocity of the ball?

If the building were twice as tall, the final velocity of the ball would remain the same. The final velocity only depends on the initial conditions (mass of the ball, initial velocity, and height of the building) and is independent of changes in the height of the building.

User JakubRi
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