TBR PHYSICS CHAPTER 1 EX. 1.15

[crazy person]

---

Members don't see this ad.

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 !

 

[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.....


Not sure if this is what youre asking for.

 

[crazy person]

Yes, this is exactly what I was asking. I realized that I had to plug the t = 2.8s back into the equation Y = V0yT + 1/2 at ^2 in order to find out how far up did the fruit go from the clip. But, did you forget to add the initial velocity in the y-direction into the equation? and this is what I was also referring to. This is what I did: Y= 26.8 (3) + 1/2(-10)(3)^2 and I got 36m. And from there, I understood how to get the flight time that it took in 350m. And, now we can add up the total flight time which is 2.8s + 8.5 s which give us 11 seconds. Finally, we can plug it back into the equation: d = Vx t which gives us 253m in order to find the range. Am I correct? I hope the initial velocity in the y-direction that I plugged back into the equation wasn't a lucky guess, but I'm more confident now. Please let me know.

Yes, I was getting confused with the different heights and forgetting to find out how far up the fruit went from the clip as you pointed out. Thank you !

 

[LabratABQ]

Yes, this is exactly what I was asking. I realized that I had to plug the t = 2.8s back into the equation Y = V0yT + 1/2 at ^2 in order to find out how far up did the fruit go from the clip. But, did you forget to add the initial velocity in the y-direction into the equation? and this is what I was also referring to. This is what I did: Y= 26.8 (3) + 1/2(-10)(3)^2 and I got 36m. And from there, I understood how to get the flight time that it took in 350m. And, now we can add up the total flight time which is 2.8s + 8.5 s which give us 11 seconds. Finally, we can plug it back into the equation: d = Vx t which gives us 253m in order to find the range. Am I correct? I hope the initial velocity in the y-direction that I plugged back into the equation wasn't a lucky guess, but I'm more confident now. Please let me know.

Yes, I was getting confused with the different heights and forgetting to find out how far up the fruit went from the clip as you pointed out. Thank you !

Yes! you got it right. WOOT

About the Voy. You can, but i was too lazy and i dont know to do more calculation then i have to. So think about it for a sec.

A. What I did: Y = 1/2 at^2 = 0.5 (10) (2.8)^2 = 39 m

B. With Voy: Y = Voy(t) + 1/2 at^2 = 27(2.8) + 0.5 (-10)(2.8)^2 = 36 m.
Reason why I used -10m/s^2 for a in method B because remember the direction of the acceleration. I assigned downwards direction as negative and upwards as positive.

You can do either one, but for A, I treated as a free fall problem, save me time and chances that I will mess up my math. lol Whatever is easier for you, you should do it. 🙂