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(II) A shot-putter throws the shot (mass $=7.3 \mathrm{kg}$ ) with aninitial spced of 14.4 $\mathrm{m} / \mathrm{s}$ at a $34.0^{\circ}$ angle to the borizontal.Calculate the horizontal distance traveled by the shot if it leavesthe athlete's hand at a height of 2.10 $\mathrm{m}$ above the ground.

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22.3 $\mathrm{m}$

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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02:34

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.

02:21

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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So here we can choose the origin to be the point of release of the shot put. We're gonna choose upwards to be positive And so we can say that why initial equals zero meters. We can save me. Why initial would equal here. It would be positive. 14 0.4 sign of 34 degrees. This is equally insider of five meters per second. Ah, we're going to say that here acceleration in the direction would equal G which would equal negative 9.80 meters per second squared. And we can say that here, why final would equal negative 2.10 meters. Given that the shot put it, the shot putter throws the shop. Put 2.10 meters above the ground to its final location would be a two point negative 2.10 meters s so we can find the time of flight. We can say that why final equals y initial plus V y initial T plus 1/2 gt squared. We know that why final is going to be sorry. What initials gonna be zero. And so we can say, uh, that 1/2. Hey, why t squared green simply said 1/2 of G t squared plus V. Why initial t minus y final equals zero. Uh, we can say negative. 4.9 meters per second squared times t squared, Uh, here. Plus 8.5 meters per second times t and then plus 2.10 meters, this equal zero. As you can see, this is a quadratic equation that equals zero. So we can, uh, use t I 84 85 or 99 in order to solve for T. We're gonna choose the positive value. So here T is gonna equal 1.872 seconds and for the total distance traveled, this would simply be v X initial t plus 1/2 ace of X t squared. There isn't any acceleration in the ex direction. So this term is eliminated and this becomes 14.4 co sign of 34 degrees meters per second multiplied by 1.872 seconds and we find that the total horizontal displacement would be equal to approximately 22.3 meters. This is our final answer that it's the end of the solution. Thank you for watching

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