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alexandr402 [8]

3 years ago

14

A ball is thrown downward from the top of a 55.0 m tower with an initial speed of 11.0 m/s. Assuming negligible air resistance,

what is the speed of the ball just before hitting the ground?

1 answer:

irina1246 [14]3 years ago

8 0

I'm not positive however my guess is this.
KE=(1/2)mv^2
KE=(1/2)(55)(121)
KE=3327.5m/s^2
You might not want to use this tho.

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A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.

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Ki+Ui=Kf+Uf

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Ui=initial potential energy

Kf=final kinetic energy

Uf=final potential energy

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u=initial velocity u

v=final velocity v, which is 0

g=acceleration due to gravity

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