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Two balls are released simultaneously from the top of a tall building. Ball A is simply dropped with no initial velocity, and ball B is thrown downward with an initial velocity of 12 m/s.

What are the velocities of the two balls 1.5 seconds after they are released?

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Final answer:

After 1.5 seconds, ball A dropped from rest would have a velocity of 0 m/s, while ball B thrown downward with an initial velocity of 12 m/s would have a velocity of approximately -12.9 m/s.

Step-by-step explanation:

The velocity of ball A dropped from rest would be zero after 1.5 seconds, as it will only experience the effects of gravity. On the other hand, the velocity of ball B thrown downward with an initial velocity of 12 m/s will also be affected by gravity. It will decrease from 12 m/s due to the downward acceleration caused by gravity.

By 1.5 seconds, both balls would have traveled a distance downwards. The distance traveled by ball A can be calculated using the kinematic equation:
distance = initial velocity * time + 0.5 * acceleration * time^2

Since ball A was simply dropped, the initial velocity and acceleration in the vertical direction will be zero. Therefore, the distance traveled by ball A after 1.5 seconds will also be zero.

For ball B, the distance traveled can be calculated using the same equation, but with an initial velocity of 12 m/s and an acceleration of -9.8 m/s^2 (negative because it is acting downwards). Plugging in the values:
distance = 12 * 1.5 + 0.5 * (-9.8) * (1.5)^2

Simplifying the equation, we find that ball B has traveled a distance of approximately 15.225 meters downwards after 1.5 seconds.

Therefore, the velocity of ball A after 1.5 seconds is 0 m/s, and the velocity of ball B after 1.5 seconds is approximately -12.9 m/s.

User Nick Bray
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