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The diff divers of Acapulco push off horivontally from rock platforms about 35 $\mathrm{m}$ abovethe water, but they must clear rocky outcrops at water level that extend out into the water 5.0 $\mathrm{m}$ from the base of the cliff directly under their launch point.See Fig. $57 .$ What minimum pushoff spced is necessary to clear the rocks? How long are they in the air?

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2.7 seconds, $1.9 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

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In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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we're gonna choose downwards to be the positive y direction. And so the origin is at the point from which the divers pushed off the cliff s so we can say that the time of flight would be equal to Y final. Whether we can use the equation, my final legal, my initial plus V y initial T plus 1/2 gt squared again. We're choosing downwards to be positive. So here, gravity is going to be is gonna be positive as well. We're going to say that my initial is zero ve y initial zero. Therefore, tea is simply gonna be the square root of two y final divided by G. This would be a square root of two times 35 meters, divided by 9.8 meters per second squared, and this is equaling 2.7 seconds. So that should be the time of flight. And then, if you want to find the horizontal speed, we can say that Delta X equals B X Times T. There's no acceleration in the ex direction, so we don't have to account for this. And so we can say that V acts initial, which would again equals the X final would equal Delta X divided by t, and this would equal 5.0 meters divided by 2.7 seconds. This is equaling 1.9 meters per second. This would be your velocity in the ex direction. That is the end of the solution. Thank you for watching.

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