Projectile motion apple throw problem

[phee]

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.

 

[ehild]

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

ehild

 

[zspdztzx]

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.

:)

 

[Rayquesto]

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, let's 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 :)

 

[phee]

Urgh, How are you guys getting 1.23 I don't 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

 

[gneill]

Urgh, How are you guys getting 1.23 I don't 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?

 

[phee]

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

 

[gneill]

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.

 

[phee]

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

 

[gneill]

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?

 

[phee]

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

 

[gneill]

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?

 

[phee]

Does your equation reflect those details?

I don't think so, should it be -4 m dor dy and -9.8 m/s ^ 2 for ay?

 

[gneill]

I don't think so, should it be -4 m dor dy and -9.8 m/s ^ 2 for ay?

Yup!

 

[phee]

Yup!

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.

 

[gneill]

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?

 

[phee]

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, don't know how to do on the forum:

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

t = 1.27s

 

[gneill]

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, don't 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.

 

[phee]

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. :(

 

[gneill]

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
at^2 + bt + c = 0
solve for t.

 

[phee]

Stupid post

 

[gneill]

Two of the ts would cancel out leaving one alone, ugh it slipped by me.

?

 

[phee]

?

Im so lost, so the equation I have been using is wrong?

 

[gneill]

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.

 

[Rayquesto]

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

 

[mackenzieb]

I also am having troubles with this problem.
I keep getting answers similar to 1.07 s which i know is not correct.
This is how i have been doing the problem:
dv=viyXt+1/2(9.8m/s)^2(t)^2
-4m=(2.86m/s)(t)+1/2(9.8m/s)^2(t)^2
t=square root of: t (-4) / -9.8+2.86
=1.07 s

Can someone please advise me as to what I am doing wrong? I have been working at this question for a while now.

 

[gneill]

I also am having troubles with this problem.
I keep getting answers similar to 1.07 s which i know is not correct.
This is how i have been doing the problem:
dv=viyXt+1/2(9.8m/s)^2(t)^2
-4m=(2.86m/s)(t)+1/2(9.8m/s)^2(t)^2

The acceleration due to gravity should be in the downward direction!

 

[mackenzieb]

--------------------------------------------------------------------------------

I am still getting 1.07 s as my answers even after putting the acceleration due to gravity as a negative. I keep getting answers similar to 1.07 s which i know is not correct.
This is how i have been doing the problem:
dv=viyXt+1/2(-9.8m/s)^2(t)^2
-4m=(2.86m/s)(t)+1/2(-9.8m/s)^2(t)^2
t=square root of: 2(-4) / -9.8+2.86
=1.07 s

I am still unsure of where I am going wrong can someone please clarify

 

[gneill]

Can you explain how you're going from the equation of motion to the t = square root... part? I don't recognize the application of the quadratic equation there.

Also, be careful with your units for the acceleration g. You've written (-9.8m/s)^2, which is not correct. First, the units of g are m/s². Second, g shouldn't be squared here, only the time is squared on the acceleration term of the equation of motion.

 

[mackenzieb]

I used this reply as reference "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"

and interpretted it into my equation...
-2.86 +/- sqrt 2.86^2-2(-4.906)(4) / 2(-4.905)
-2.86 - sqrt 8.1796+39.248 / -9.81
=-1.43

I keep finding that each time I do the equation I get a different answer. I think I am having the most trouble finding the proper equation

 

[gneill]

Your equation looks okay. You've written it several times. Here it is again:
-4m = (2.86 m/s) t + \frac{1}{2} (-9.8 m/s^2) t^2
Collecting onto one side:
\frac{1}{2} (9.8 m/s^2) t^2 - (2.86 m/s) t - 4m = 0

Can you identify and isolate the coefficients for plugging into the quadratic formula? a=?, b=?, c=?

 

[mackenzieb]

if i am correct..
a=-9.8 (not sure if this is supposed to be -, but i believe so)
b=2.86
c=-4

 

[gneill]

Take a closer look at the equation in its 'one sided' form. What exactly is multiplying t²? What sign does the t term have?

 

[mackenzieb]

there is no negative sign, therefore 9.8 should remain positive.
making the variables:
a=9.8
b=2.86
c=-4

 

[gneill]

there is no negative sign, therefore 9.8 should remain positive.
making the variables:
a=9.8
b=2.86
c=-4

What happened to the 1/2 also multiplying the t² term? And the t term is negative!

 

[mackenzieb]

does that make the a term -4.9? (9.8 divided by 1/2)

 

[gneill]

does that make the a term -4.9? (9.8 divided by 1/2)

Nope. The t² term is positive. So it's 4.9 (actually, you should probably carry an extra decimal place or so through these calculations in order to avoid rounding errors creeping into the results).

The t term is negative.

 

[mackenzieb]

okay so to make sure I am onthe right track then...
a=4.906
b=2.86
c=-4

-4m=(2.86m/s)t+1/2(-9.8m/s^2)t^2
1/2(9.8m/s^2)t^2-(2.86m/s)t-4m=0
-2.86 +/- sqrt 2.86^2-2(4.906)(4)/2(-4/906)

is that the correct setup?

 

[gneill]

Not quite. The t term is negative! b = -2.86.
And the quadratic formula is
x = \frac{-b ± \sqrt{b^2 -4ac}}{2a}

See: http://en.wikipedia.org/wiki/Quadratic_equation

 

[mackenzieb]

so when i use that equation...i get

t=-2.86+/- sqrt -2.86^2-4(4.906)(-4)/2(4.906)

 

[gneill]

so when i use that equation...i get

t=-2.86+/- sqrt -2.86^2-4(4.906)(-4)/2(4.906)

Well, the algebra as written is a bit ambiguous, but that's the general idea. What result do you get?

 

[mackenzieb]

when i get to the final step...
-2.86+/-sqrt-2.86^2-4(4.906)(-4)/2(4.906)
i get -2.86+/- 2.677007231

i am unsure what to do with the -2.86+/-

 

[gneill]

Your casual mathematical grammar is making it difficult to determine where your problem lies. I can't tell what your intention is when you have open-ended square roots and dangling divisions. Please make more use of parentheses to delimit the mathematical operations. If you choose not to use Latex to write your mathematics, then at perhaps you can at least treat the square root as a function: sqrt( stuff ), with brackets to delimit the argument.

The quadratic formula provides TWO roots to a given quadratic equation. The ± indicates how to alter the given formula to produce the two roots. While both roots may satisfy the mathematical requirements of the equation, only one may satisfy the physical constraints (things like time being a positive quantity, or distance a positive value).

 

[mackenzieb]

here is my complete work so far:


Given:
d=4.0m [down]
a=9.8m/s^2 [down]
vi=0m/s (whenever something falls from a height, the v1 is always 0)

Required:
t
vf

Therefore, I use the equation
d=vi(t)+1/2a(t)^2
4m [down] = 0m/s(t)+1/2 (9.8m/s^2[down])(t^2)
4m[down]=1/2(9.8m/s^2[down])(t^2)

Because the initial velocity (vi) is 0 the equation can be simplified...

t=sqrt 4.0m[down]/1/2(9.8m/s^2 [down])
t=0.9035 s

My question is, where am I going wrong ?

 

[gneill]

There's nothing wrong other than this is not the same problem that was given in the first post of this thread! There it was stated that the object was thrown with an initial upward angle of 35°.

 

[mackenzieb]

Yes that is right, the object was thrown with an initial upward angle of 35 degrees. I am trying to use the examples in my notebook I have been provided with, and that it what is making this so difficult.

I didnt think the angle had to be applied until the second question, but i must be incorrect.

I am unsure of how to incorportate the angle into my answer (ie i don't know where it needs to be placed in the answer and how to apply it )

 

[gneill]

Yes that is right, the object was thrown with an initial upward angle of 35 degrees. I am trying to use the examples in my notebook I have been provided with, and that it what is making this so difficult.

I didnt think the angle had to be applied until the second question, but i must be incorrect.

I am unsure of how to incorportate the angle into my answer (ie i don't know where it needs to be placed in the answer and how to apply it )

The elevated launch angle means that there will be an initial velocity in the y-direction.

 

[mackenzieb]

I don't understand what you mean by that
so I do need to incorporate the angle into this question
...meaning the initial velocity is still 0 but is 35 degrees in the y-direction?
I don't know how I am supposed to place this into the equation

 

[gneill]

I don't understand what you mean by that
so I do need to incorporate the angle into this question
...meaning the initial velocity is still 0 but is 35 degrees in the y-direction?
I don't know how I am supposed to place this into the equation

The launch speed is 5 m/s as stated in the problem. The launch is at an angle to the horizontal, so that this speed will have a horizontal and a vertical component. What are they?

The initial velocity components plug into the equations of motions that pertain to the vertical and horizontal motions.

 

[MazharB]

Gneill when i use the quadratic form, and plug in the given variables i do not get 1.27 or 1.23 or anywhere near that
∆t = -2.86+/- sqrt -2.86^2-4(4.906)(-4)/2(4.906)
t=-3.8 or -1.91