TBR EX. 1.15 Chapter 1

[crazy person]

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Hi all,

Can someone please explain to me the concept behind the turbo solution. I get that instead of using the quadratric equation, we can use a short cut. I get why we are using the formula v0/g and get 26.8 for the upward flight but I don't understand why we can't add 2.8 seconds plus 2.8 seconds to get the total flight time when it reaches the ground, this way we can just plug in into the x direction equation which is x= vxt .
If they just added the other flight time using d= voyt + 1/2 at^2 then why didn't they add the initial velocity 26.8 in order to find the other time flight that it takes to get down if this is the displacement equation and velocity is giving in the y direction. I guess I'm confused why the initial velocity in the the y direction is not added to their turbo shorcut if you need the velocity in the y direction. Help please !

 

[inasensegone]

I'm not really sure what you're asking, so instead I'll explain the turbo method the way I see it being employed, and then you can tell me if that clarifies things for you.

So the apricot is launched off of a cliff. Given that the velocity in the Y direction (straight up) is given by Sinϴ Vo, we arrive at the value 26.8 [because the Vo was 35 m/s and the sin of the launch angle (50 degrees) is equal to 0.766 as given. Multiply together and you get 26.8]. OK, so far so good, sounds like you understand this much.

Now, to figure out the flight time for the first portion of the flight, we simply divide the velocity in the y direction by gravity, which gives us ~ 2.68 seconds (assuming g=10) or approximately 2.8 seconds as they round it. Notice that the first portion of the flight is not equivalent to half the flight time. This is why we can't simply double this value of 2.8 seconds. Because the projectile is launched off of a cliff, we must now consider the initial displacement in the Y direction (the 310 meter height of the cliff) and add that displacement to the distance the projectile reached at its apex after the 2.8 seconds of vertical flight.

We must estimate this displacement during the first half of the flight by the turbo method. We know that if the object were on level ground, the first half of the flight time would equal the 2nd half of the flight time. This tells us that it takes the object the same amount of time to fall to the ground, as it did to reach it's apex. Because of this we can estimate that the object is approximately ~ +40 meters at its apex. How? Well, recall the chart that was provided for a free-fall object on page 21: At 2 seconds an object will fall 20 meters, and at 3 seconds it will have fallen 45 meters. Therefore we can conclude that 2.8 seconds will be closer to 45 meters than 20 meters. We approximate 40. Add this 40 to the 310 height of the cliff and we get 350. How long does it take to fall 350 meters? Again, the chart indicates just over 8 seconds. Add that to the 2.8... let's call it 11 seconds. That means 11 seconds total flight time. Multiply this by the velocity in the Horizontal direction (Cos ϴ Vo ~ 22.5 m/s) and you get 247.5 m, but you can call that 250. The actual answer is 251 meters.

Now you might be asking how you know you can approximate that much and arrive at the correct answer. Well, look at the difference in the answers. The nearest values are 180 m and 390 m. There is plenty of room there. Also, keep in mind, this took way longer to explain than it should take you to make these approximations in your mind. You should become familiarized with the values of distance traveled by an object in free fall. You should also becomes familiar with your trig values, your projectile equations, and all the other shortcuts involved in this question. This will make arriving at the correct answer a ~30 second process

I hope that this answered your question. Let me know if it doesn't.

 

[crazy person]

Yes it did ! Thank you. So pretty much, I would just add 40 meters to the height of the cliff after looking at the chart and then I would approximate again to see how long it takes to fall 350 meters which is 8 seconds according to the chart. And after I have done all this, then I can multiply by the velocity in the Horizontal direction. So the key here is the chart that is provided in order to to a quick free fall estimation, right? So, this means I will have to memorize the chart, correct?

What do you think about this following method in order to get the same solution. Since we already know the time it takes to go up is 26.8 seconds, now we could use : t = square root 2d/g which derives from d= vt + 1/2 gt ^2 (and this is why I was asking if this formula is use, then why wouldn't we include the initial velocity 26.8 into v ), because once we plug t = square root 2(310)/ 10 which equals square root 62; we can estimate the answer to be 8. So, now we can add that to 2.8 which give us around 11 . And now we can multiply this by the velocity in the Horizontal direction (Cos ϴ Vo ~ 22.5 m/s) which give us the same answer 247.5 m. Let me know what you think.

 

[crazy person]

and YES !!!! your explanation was really helpful !!! Thank you !

Let me know if you think the other method can also be used to solve the problem as well =)

 

[inasensegone]

No problem! The chart really isn't that hard to remember, especially if you practice enough projectile questions (which you should be doing anyway). Also, recall TBR's side note in regards to the chart: The distance traveled in time t, will be 4 times as far in time 2t. For example, @ 2 seconds the object falls 20 meters. At 4 seconds, the object falls 80 meters. 20 times 4 equals 80. Same thing for 4 seconds and 8 seconds. At 4 seconds, distance = 80 meters. At 8 seconds, distance = 320 meters.

In regards to the 2nd paragraph you posted, I'm not really sure I follow your equation of time = square root of 2d/g? I think this means that you are assuming the object is in free fall from the 310 meter mark, and then calculating time and adding it to the 2.8 seconds. This doesn't work though because in the downward trip of the projectile's pathway (past the apex) it already has some velocity in the y direction by the time it gets to d = 310 meters (this velocity is equivalent to the launch velocity in the y direction). Therefore you can't really calculate time from this point on, unless you factor that velocity in. This is exactly where the quadratic equation came into play with TBR's explanation, because the equation is Dy = Vy+1/2gt^2 which becomes -310 = 26.8t + 1/2(-10)t^2.

Let me know if that answers your question.

 

[crazy person]

I actually understand it both ways. so Thank you !
This is what I meant to tell you in order to find the answer without doing the turbo. Another poster explained it a lot better than I did: LabratABQ

Are you asking why the total time in flight doesnt equal to 2.8+2.8 = 5.6 s?

If you're then you are not absolutely wrong, you have the total time in flight concept correct, however that is when the object took off and landed at the same attitude, in another word the initial height and the final height must be the same.

In this problem, the initial height is 310m higher than the final height, the fruit is being thrown off the cliff from a higher grown. So if you think about it:

t=2.8s is the time that it takes to reach apex, from where the dude was standing. You do not know how far up did the fruit go from the clip, so you must plug the t = 2.8s back into y = 1/2 at^2 to find the distance that the fruit travels in that 2.8s from the clip. You will get around 36m, so i round it up to 40m,

now for the Y component, we need to find how long it took for the fruit to travel the total distance of 310 m + 40 m = 350 m.
Y=1/2at^2
350 = 0.5 (10) t^2
t ^2 = 70
we know that 8^2 = 64, and 9^2 = 81, therefore 8^2 < t^2<81
t= ~ 8.5 s, the time it takes to travel the 350m in the vertical direction is 8.5s


Now, since we have established that the initial and the final height arent the same, thus the total time in flight is = 2.8 s + 8.5 s = 11 sec.
From there you can plug it back into the X = Vox(t) + 1/2 at^2 and solve for the range that the fruit traveled, knowing that in the x component, a = 0 or you can do the estimate thing that they showed you in the book....1s = 5m, 2s = 20m.....

 

[crazy person]

Thank you !!!🙂

and I know this is a long post but I wanted to share with you the other way. But I'm sure you have it down either way.

 

[inasensegone]

No problem! The difference between his post and yours, was that you said time = sq.rt of (2*310)/10... leading you to take the square root of 62. LabratABQ did this correctly in that you would actually be doing time = sq.rt of (2*350)/10... leading you to take the square root of 70. A minor math mistake on your part so no biggie...

Either method works, I was just explaining the turbo method to you since that is what you were asking about with regards to TBR's method.