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Problem 26 Medium Difficulty

In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in Figure 3.10 and has a range of 23 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?

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Sharieleen A.

October 23, 2020

Finally, the answer I needed, thanks Griffin G.

Catherine A.

October 23, 2020

That was not easy, glad this was able to help

Kaitlyn H.

October 15, 2020

At the end, why did you do V2^2/v1^1?

Top Physics 101 Mechanics Educators

Christina Krawiec

Rutgers, The State University of New Jersey

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Video Transcript

So the question states that a projectile is shot at some angle theta at a speed V and travels 23 meters. And we're trying to find how far this project I will go. If we doubled the speed it was launched it but kept the angle the same. So the first thing we need to do is separate are, um, velocity vector into components. So we know this is going to be equal to be co sign data, and this will be equal to be signed data. And we get this because scientist Aita is opposite over hype on news. So opposite is, um, our visa by or this vertical velocity vector and are high pot news is V so that we could just solve for be supply. Same thing goes for Vico Sant data in height. Find it. Um, so now that we know this, we know that the range of the project I was going to be equal to the frosty times, the time evil, the delta tax and the velocity in the horizontal direction is going to be equal to the velocity Times co sign data and then this will be multiplied by the time which will give us our tell tax. So the main thing we need to do here is so for T in terms of Fada and V And to do this we can use our kinetic equations which state that the change in displacement in this case in the Y direction is equal to the initial velocity times a time plus for 1/2 times the acceleration time to the time squared. And so the we're told in the problem that the projectile starts on the ground and it ends on the ground. So its displacement is zero. We know its initial vertical velocity is the sign data. I don't know what tea is, and we can just leave acceleration as a We know it's going to be negative 9.8. But honestly, that is a and then t squared. And so from this we confected at a T. So get he signed Data plus one have a t all this most played by T and it's equal to zero. And so when we saw for tea will find that he is either equal to zero or a T is equal to negative two. The sign data over a So we can ignore the route that says that T is equal to zero because this doesn't make any sense in our problem. So we can use this route and plug it back into our equation up here. And when we do this, we find that the range is equal to, uh minus two times the velocity squared times sign. They, uh, Times co sign data all over the acceleration. So if we tailor this to our problem, let's say, uh, for the, um, the situation where the projectile only goes 23 meters, we'll call this Delta X one. And we'll also called this velocity V one. Um, and we can say this is also the one. It's still just make. It would be to be one anyways, So if we say Delta X one is going to be equal Teoh minus two times the velocity of the one square times signed data Times co sign data all over the acceleration. And if we look at the second situation, we can say that the, um, the displacement. So we'll call it Delta X two, And I'm actually gonna change this to, instead of to be one. I'll just make this be to so Delta X to the range of the velocity. The range of the projectile here is equal to minus two times feet too. Squared times signed data co sign data because the angles the same in both situations divided by a And so we confined the ratio between these two. So I'm gonna take dealt, uh, x two and divided by Delta X one. And when we do this will end up with the velocity of the second projectile squared, divided by the velocity of the first projectile squared. And this gives us a relationship where we confined. Um, how far this project I was gonna dio, given that the velocity is two times the original projectiles, boss T. So this V two is actually the one times two. So when we plugged that in, we see that Delta X two over Delta X one is equal to four v one squared over the one squared in This cancels and we see that Delta X two is equal to four times Delta x one, which means that the range of the second projectile is four times the range of the first projectile, which means it's going to be four times 23 meters, which is equal to 90 to meters. And that's the answer

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John D. Cutnell, Kenneth W. Johnson

Physics

Elyse Gonzalez

Christina Krawiec

Andy Chen

Marshall Styczinski

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

Physics

Elyse Gonzalez

Christina Krawiec

Andy Chen

Marshall Styczinski

John D. Cutnell, Kenneth W. JohnsonPhysics

Elyse Gonzalez

Christina Krawiec

Andy Chen

Marshall Styczinski

John D. Cutnell, Kenneth W. Johnson

Physics