Projectile motion apple throw problem

  • Thread starter phee
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  • #1
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Homework Statement



A child sitting in a tree throws his apple core from where he is perched (4.0m high) with a velocity of 5.0 m/s [35 degrees above horiztontal], and it hits the ground right next to his friend.

A) How long is it before the apple core hits the ground?

B) How far from the base of the tree will the apple core land?

C) What is the velocity of the apple core on impact?


Homework Equations



part A) I used dy= Viy x t + 1/2(ay)(t)^2

part B) I used dx = (Vx)(t)

part C) I used Vf^2 = Viy^2 + 2ay(dy)

Then Pythagorean theorem to find out the final velocity and tan function to find the angle


The Attempt at a Solution



My answers were as follows:

A) 1.27 second

B) 5.2m

C) 9.9 m/s @ 65.5 degrees below horizontal


Any feedback is greatly appreciated.
 

Answers and Replies

  • #2
ehild
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It is all right, but there are rounding errors. Use more significant digits during calculations.

ehild
 
  • #3
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Homework Statement



A child sitting in a tree throws his apple core from where he is perched (4.0m high) with a velocity of 5.0 m/s [35 degrees above horiztontal], and it hits the ground right next to his friend.

A) How long is it before the apple core hits the ground?

B) How far from the base of the tree will the apple core land?

C) What is the velocity of the apple core on impact?


Homework Equations



part A) I used dy= Viy x t + 1/2(ay)(t)^2

part B) I used dx = (Vx)(t)

part C) I used Vf^2 = Viy^2 + 2ay(dy)

Then Pythagorean theorem to find out the final velocity and tan function to find the angle


The Attempt at a Solution



My answers were as follows:

A) 1.27 second

B) 5.2m

C) 9.9 m/s @ 65.5 degrees below horizontal


Any feedback is greatly appreciated.

Your rationale is correct! However, I have a few minor suggestions -

1. For A), B), and C), I agree with ehild, there are rounding problems. My results for A) are 1.23s (for g=-10) or 1.24s (for g=-9.8).

2. For C), why not using Vyfinal=Vyinitial-g*t? I think this makes calculation easier, as you avoid doing sq root.

:)
 
  • #4
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Oh good! I know how to do this. Sounds like a problem I had trouble with, but I remember a really good way to figure it out. So, as long as you know that dy=vi,yt + 1/2gt^2 then you could figure out time. since the apple core falls 4 meters below the point of full quadratic motion (basically 4 meters below the y point in which it returns),

-4meters= vi,yt +1/2gt^2

vi,y=5m/s(sin35) and plug that in and use the quadratic formula to find time and use common sense to determine which time makes more sense. and use that time to find distance which is easy.

As for the third question, lets see v^2=v0^2 + 2adx and v=v0 + gt and v^2=v0^2 + 2gdy
So, on impact, v^2=(5.0m/s)^2 +2g(-4meters)

Figure out velocity from there :)
 
  • #5
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Urgh, How are you guys getting 1.23 I dont know what I am doing wrong??

for time I use

4 = 2.86m/s x t + 1/2(9.8 m/s^2) t^2

I keep getting 1.27s

Can anyone point out where I am going wrong? This is killing me
 
  • #6
gneill
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Urgh, How are you guys getting 1.23 I dont know what I am doing wrong??

for time I use

4 = 2.86m/s x t + 1/2(9.8 m/s^2) t^2

I keep getting 1.27s

Can anyone point out where I am going wrong? This is killing me

Can you explain the terms in your equation? How do the signs of the terms in your equation relate to your chosen coordinate system?
 
  • #7
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Can you explain the terms in your equation? How do the signs of the terms in your equation relate to your chosen coordinate system?

dy =vi,y x t + 1/2 a,y x t^2
 
  • #8
gneill
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dy =vi,y x t + 1/2 a,y x t^2

That's not a clear explanation. When you stuck in the numbers you wrote:

4 = 2.86m/s x t + 1/2(9.8 m/s^2) t^2

Explain your choice of signs for each of the terms.
 
  • #9
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That's not a clear explanation. When you stuck in the numbers you wrote:

4 = 2.86m/s x t + 1/2(9.8 m/s^2) t^2

Explain your choice of signs for each of the terms.

I got the 4 from the height (dy) of the child.

2.86m/s from 5m/s x Sin35

t = unsolved for

9.8 m/s^2 = acceleration due to gravity
 
  • #10
gneill
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I got the 4 from the height (dy) of the child.

2.86m/s from 5m/s x Sin35

t = unsolved for

9.8 m/s^2 = acceleration due to gravity

So... The apple core is going to land 4 meters higher than the child in the tree, and is accelerated upwards at a rate of 9.8m/s2?
 
  • #11
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So... The apple core is going to land 4 meters higher than the child in the tree, and is accelerated upwards at a rate of 9.8m/s2?

No it will land 4 m below the child on the ground, and it is thrown at an angle of 35 degrees at 5m/s
 
  • #12
gneill
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No it will land 4 m below the child on the ground, and it is thrown at an angle of 35 degrees at 5m/s

Does your equation reflect those details?
 
  • #13
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Does your equation reflect those details?

I dont think so, should it be -4 m dor dy and -9.8 m/s ^ 2 for ay?
 
  • #14
gneill
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I dont think so, should it be -4 m dor dy and -9.8 m/s ^ 2 for ay?

Yup! :smile:
 
  • #15
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Yup! :smile:

That still gives me 1.27s though. I still don't know what I am doing wrong to keep being .3s off the times that others are getting.
 
  • #16
gneill
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That still gives me 1.27s though. I still don't know what I am doing wrong to keep being .3s off the times that others are getting.

Can you show your calculations?
 
  • #17
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Can you show your calculations?

My original calculations were:

4 m = 2.86 m/s x t + 1/2 (-9.8m/s ^2) t^2

4 m = 1/2(-9.8m/s^2) t^2

This is in square root, dont know how to do on the forum:

t = 2(4m) / (9.8m/s^2)

t = 1.27s
 
  • #18
gneill
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My original calculations were:

4 m = 2.86 m/s x t + 1/2 (-9.8m/s ^2) t^2

You've still assumed that the landing point is above the launch point.

4 m = 1/2(-9.8m/s^2) t^2
What happened to the velocity term (t) ?
This is in square root, dont know how to do on the forum:

t = 2(4m) / (9.8m/s^2)

t = 1.27s

You need to use the full quadratic formula here. There's a "t" as well as a "t2" term to deal with.
 
  • #19
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You've still assumed that the landing point is above the launch point.


What happened to the velocity term (t) ?


You need to use the full quadratic formula here. There's a "t" as well as a "t2" term to deal with.

The first t is negated out because its unknown which takes away the 2.86 m/s right?

I will redo the equation now with the negatives in what I believe is the way to do it.

-4m = 2.86 m/s x t + 1/2 (-9.8 m/s ^2) t^2

-4m = 1/2 (-9.8 m/s ^ 2) t^2

t^2 = -4m / 1/2(9.8 m/s^2)

And from square rooting the answer of that I get error on my calculator. :(
 
  • #20
gneill
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The first t is negated out because its unknown which takes away the 2.86 m/s right?

No. The term must stay. The t is the unknown that you're solving for.

The equation is of the form
[tex]at^2 + bt + c = 0[/tex]
solve for t.
 
  • #21
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Stupid post
 
  • #22
gneill
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Two of the ts would cancel out leaving one alone, ugh it slipped by me.

:confused:???
 
  • #23
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:confused:???

Im so lost, so the equation I have been using is wrong?
 
  • #24
gneill
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Im so lost, so the equation I have been using is wrong?

It's not wrong. Once you've got the signs of the terms straightened out it's fine. But you can't just drop terms that are part of it.

Rearrange the equation into the form I showed and plug the coefficients into the quadratic formula to solve for t.
 
  • #25
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hey Phee, go back to my post if you want. I explained that you should use the quadratic formula. as long as 0=ax^2 + bx + c then you can solve for x using a particular quadratic formula

so if 0 = 1/2(-9.81m/s^2)t^2 + 2.86 m/s t + 4m you can solve for t using

(-b +/- sqrtb^2-2ac)/2a=t

a=-4.905
b=2.86
c=4

since at^2 + bt + c=0
 

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