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At scrve, a tennis player aims to hit the ball horizontally.What minimum speed is required for the ball to clear the$0.90-$ m-high net about 15.0 $\mathrm{m}$ from the server if the ball is "launched" from a height of 2.50 $\mathrm{m}$ ? Where will the ball land if it just clears the net (and will it be "good" in thesense that it lands within 7.0 $\mathrm{m}$ of the net)? How long will it be in the air? See Fig. $58 .$

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$18.8 \mathrm{m}$

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

Cornell University

Rutgers, The State University of New Jersey

University of Winnipeg

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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.

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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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Our question says that serve a tennis player aims to hit the ball horizontally. What minimum speed is required for the ball to clear a 0.9 meter high net about 15 meters from the serve or from the server? If the ball is launched from a height of 2.5 meters, where will the ball land if it is to clear the net, and will it be good in the sense that it lands within this within seven meters of the net on the other side? How long will it be in the air Sea figure 58 kind of set. Something lays out this problem and I also wrote down here what we were given, which is that the distance it's going to travel to get to the net call access 15 meters. Why is the height of the net 0.9 meters? Why not is the initial height from the server 2.5 meters? We're assuming all the velocity is in the X direction, right, so there's no initial velocity in the white direction because they launch the ball horizontally. Okay, so first we want to figure out how long it's going to take for that ball to reach the net. And in order to do this, we're going to make use of the equation that we've been using. Why is equal to why not? Plus the of Why the initial velocity in the white direction times the time, plus 1/2 the acceleration in the wind direction times The Times Square. Okay, well, let's go ahead and plug in what we know for our situation here in order to find the time it's going to take for it to reach the net while at the net. Why 0.9 meters, right? Okay, why not is 2.5 meters, We said all the velocity is in the X direction since it's launched horizontally. So vo zero y zero. So this is going to be zero and then minus here because the acceleration in the white direction is gravity in the negative direction. So we have the minus side pulled out front. Well, plug in gravity, which is 9.8 meters per second squared. This is multiplied by t squared. But this time here is the time it takes for it to get to the nets. Will call this T's of in for net squared. Okay, so now let's solve for the time it takes for it to get to the Net. T seven. This is going to be equal to square root since t's is squared. Sand the square root. Yeah, you have two right from the 1/2 and that's going to be multiplied by 0.9 minus 2.5 meters, which is minus 1.6 meters, divided by the negative 9.8 meters per second squared from gravity. Oh, okay. So phew. Playing all those values into a calculator, you find that this comes out to be 0.571 seconds. That's how long it's going to take for it to get to the net. Okay, so then, to find the velocity next direction that's required for it to clear the net, we can make use of the distance formula, which says that distance and the extraction is equal to the velocity and the ex direction times the time it travels. We want to find velocity, right? So the required velocity next direction is going to be the distance to the net. We'll call that X right, which we call X divided by the time it takes to get to the net, which is t seven. So this is going to be equal to 15 divided by 0.571 Mrs Meters per second. Flint this into a calculator and you find that they've lost in the extraction is cool. 26 no, same size of the two. 26.3 meters per second. Okay, so now that we know the velocity in the ex direction, let's solve the total time because we want to figure out if it does clear the net, right. If you just clear the net, how how far on the other side of the net is it going to go? We want to figure out if it's going to land within the safe zone. You wanna land within seven meters on the other side of the net? Okay, so let's go back and use the original equation that we had right? Which is I'm going to note here with this Green Star this original equation. Except this time we're going to assume that the ball goes all the way to the ground because it's reached the other side of the net. Let's get a new page here. So Okay. Now, if the ball reaches the ground, our new why is zero, right? Okay. Why not? It's still 2.5 meters. That's still the height of the server. Close. Um, good velocity in the white direction. Still zero plus 1/2. And then you have gravity, which is nine minus 9 20 meters per second squared. And then this Here is the total time. The balls traveling. Okay, so this he were going to notice. Thi said big T for total square. Okay, so I want to get in solving for times, figure out how long it's in the air. You're gonna have square root. Okay. Two from the 1/2 multiplied by minus 2.5 meters. Zero minus 2.5 is just minus 2.5. Divided by the negative. 9.8 meters per seconds. Squared from gravity. Okay, look, all of that into your calculator. You find that the total time the ball travels is 0.714 seconds. Okay, Now that we know the total time, the ball travels and we know that the ball is going to clear the net. We confined the total distance. It travels in the ex direction. We'll call this except tea for total distance. This is going to be the velocity in the extraction times. The total time the ball travels Well, if you go back to the previous page, we found the velocity in the ex direction, Um to be 26.3 meters per second. So Well plugged that in. So this is equal to 26.3 meters per second. Multiplied by 0.714 seconds seconds. Cancel and you're left with meters. Okay, Mrs. Equal to 18.8 meters. Okay, well, we were told the safe zone is within eight meters of erm Sorry. The safe zones with it. The good zone is within seven meters of the other side of the net. So the total distance of the good zone is within 22 meters. So if it's within 15 which is the distance to the net, if it's within 15 to 22 because 15 for seven is 22 in that range, it's good 18 point ate it. Meters is within that range, so we'll just write that it's good box in solution

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