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Problem 14 Medium Difficulty

As a tennis ball is struck, it departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket?

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Serina C.

February 24, 2021

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Video Transcript

eso in this video, We're going to look a editor you from or 14. So we know a test ball gets struck directly wars only. So we don't have to account for any up motion, which is always nice when doing projectile problems on. The first thing you should always do is can draw pictures. So we have draw the ground here, and this is where it gets struck. And then it's gonna go down gravity. And we're ignoring a resistance and everything in this problem. So, yeah, only only worry about, really what the question gives you and gravity. So that's right. This in terms of their components. So we know the initial velocity in the X is 28 0.0 meters per second. We know the initial acceleration in the axe is zero meters per second squared Should be well, the only reason it would decelerate is for, um, resistance. Anyways, um, and then so it travels an ex distance, horizontal distances says that starting with the axe of 19.6 meters. Okay, so there you go. That's everything to the axe. And now we have to do. Why so? Because this doesn't get hit up or down. At a certain degree, we know that the initial velocity in the UAE direction is zero meters per second. We know. Is it gonna accelerate in my direction? Yeah, gravity. Right. So acceleration. And the why is gravity is 9.80 I think the textbook said for this 9.0 Mitch per second squared and word of looking for this. Why we don't know how high above the ground. We don't know this distance here. Where this bull about it. That's what we want to find in this question. Right? So let's try and find the time of the flight over the ball because that'll have us right. So we know that the distance that it travels and axe is equal to V the initial time times you're in sort of the initial velocity times time plus 1/2 a T squared Look. Okay, so there you go. That's that's a formula. We have the initial velocity, we have the acceleration and we can find time. So time from your engines problem you get T is equal to the total distance Traveled, divided by initial velocity. So that's 19.6 meters divided by 28 years presented. Yeah, so this is equal to 0.700 seconds. And that's how long balls and therefore so now that we know how long the balls in the air for we can, um, weaken the time isn't in the extraction or why direction once you get, uh, I don't know how many dimensions maybe time to go on another dimension. But, um, time is the same in exit as it is for why in this question, Uh, so now we can do the same thing for what we want to find how high this ball is above the ground so we can use the same form. Why, as this one. So we can use this one for Why? So why equals V I t. Plus 1/2 a T squared? But this first term V I t will end up going to zero just because initial velocity and the Y zero right right there. So, uh, yeah, it's just cancel. So why equals 1/2 a T spurt and that this is the distance about this is perfect. This is exactly what we're looking for. This why is how high above the ground that the tennis ball get struck out. So we're gonna go. Let's just put our numbers. So 1/2 times 9.81 or zero meters per second. Squared times 0.700 seconds. Squared seconds squared. Notice how When we square seconds, you get second squared units. Cancel. Perfect. So this is gonna leave us with the final around the space here. Why equals, uh, 2.401 meters. Okay, so that's exactly were just right. How we're just right at the components of it. We want to find how long the balls in the air, because time and lower level physics isn't on each component X y z. So time is the same in next in line the balls on this f for the same amount of time. So we can use this formula to determine how high the bowl is above the ground. So I hope this video was a useful thing for watching and good luck on our street problems.

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Watch More Solved Questions in Chapter 3

Video Transcript

John D. Cutnell, Kenneth W. Johnson

Physics

Elyse Gonzalez

Andy Chen

Marshall Styczinski

Jared Enns

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics

Elyse Gonzalez

Andy Chen

Marshall Styczinski

Jared Enns

John D. Cutnell, Kenneth W. JohnsonPhysics

Elyse Gonzalez

Andy Chen

Marshall Styczinski

Jared Enns

John D. Cutnell, Kenneth W. Johnson

Physics