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GrogVix [38]
3 years ago
8

A ball is thrown straight up from the ground with an unknown velocity. It reaches its highest point after 5.5 s. With what veloc

ity did it leave the ground?
Physics
1 answer:
Maslowich3 years ago
5 0
V = u + at
0 = u -9.8 x 5.5
u = 9.8 x 5.5 = 53.9 m/s
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Answer:

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Acceleration vectors point toward the center of the circle.

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Explanation:

(THESE ARE NOT MY WORDS BTW)

1) Acceleration and velocity are vectorial quantities, which means they have magnitude and direction.

2) In a circular motion velocity direction changes all the time, which means that it is accelerated.

3) In a uniform circular motiion, the velocity changes in a constant value. This is the rate of change of velocity, which is the magnitude of the acceleration, is constant (uniform).

4) The velocity is perpendicular to the path, i.e. the circle. You can see it if you think that if the object stopped changing the direction, then the object would follow a straight path (as per inertia principle). That is why this velocity is called tangential velocity (to differentiate it of the angular velocity).

This is what the option C says "Velocity vectors are always perpendicular to the circle". Then this is true.

5) The constant change of direction in a circular path, means that the object is been pushed, accelerated, toward the center of a circle. This is, all the time the object in motion tries to follow the perpendicular path but a push (a force) directed to the center of the circle changes its direction. Such force accelerates the object toward the center of the circle. So, the acceleration vectors point toward the center of the circle, which is what the option D says. So, this is also true.

6) Since the motion is uniform, the magnitude or length of the velocity vectors are always the same, are constant. So, the option E. is also true.

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