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I'm not really sure what equation to use or how to approach.

- Thread starter louis676
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- #1

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I'm not really sure what equation to use or how to approach.

- #2

berkeman

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Welcome to the PF.

I'm not really sure what equation to use or how to approach.

Start by listing the equation for the vertical position of an object as a function of time y(t), based on the initial y position, the initial vertical velocity, and the acceleration of gravity.

Then write the two equations for the vertical motion of the two objects, given their initial conditions. Then find the time t when their y positions are equal.

Please show us your work along those lines...

- #3

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Basically this is what I did, but I'm pretty sure is incorrect. I don't really understand what you mean by "vertical position of an object as a function of time y(t), based on the initial y position," because my physic teacher never explain this to us. I know what is v-t and x-t graph, but when integrate with vertical or horizontal literal, I'm pretty much lost. My physic teacher only give us question and go over it when we meet, but he never really give us a exact explanation of how things work.

So please help me out.

Vertical position for falling object

Δx = V

Δx = ?

V

a = -9.8 m/s

Δt = ?

Δx = V

Δx = (0m/s)(Δt) + .5(-9.8m/s

Δx = 0 m/s + -4.9 m/s ((Δt)

Δx = -4.9 m/s

4.9 m/s

2.21 s = t

Vertical position of object throwing up

Δx = V

Δx = ?

V

a = -9.8 m/s

Δt = ?

Δx = V

Δx = 28 m/s (Δt) + .5(-9.8 m/s

- #4

berkeman

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So please help me out.

Vertical position for falling object

Δx = V_{o}Δt + .5a(Δt)^{2}

Δx = ?

V_{o}= 0 m/s

a = -9.8 m/s^{2}

Δt = ?

Δx = V_{o}Δt + .5a(Δt)^{2}

Δx = (0m/s)(Δt) + .5(-9.8m/s^{2})(Δt)^{2}

Δx = 0 m/s + -4.9 m/s ((Δt)^{2})

Δx = -4.9 m/s^{2}(Δt)^{2}

4.9 m/s^{2}= (Δt)^{2}

2.21 s = t

Vertical position of object throwing up

Δx = V_{o}Δt + .5a(Δt)^{2}

Δx = ?

V_{o}= 28 m/s

a = -9.8 m/s^{2}

Δt = ?

Δx = V_{o}Δt + .5a(Δt)^{2}

Δx = 28 m/s (Δt) + .5(-9.8 m/s^{2) (Δt)2 Δx = 28 m/s (Δt) - 4.9m/s2 (Δt)2 I don't know what to do here, cause I can't use quadratic equation to solve this. I believe my approach is already wrong.}

- #5

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After following your instruction, this is what I did. I pretty sure what I did was wrong though.

Can you simplify how to work on this? I'm just not getting it lol.

X =?

X

Δt = ?

V

A = -9.8 m/s

X (Δt) = X

X (Δt) =79 m + 28/ms Δt + .5(-9.8 m/s

X (Δt) =79 m + 28/ms Δt - 4.9 m/s

79 m (Δt) = 28 m/s Δt - 4.9 m/s

Ax

-4.9 + 28 + 79 =0

X = -b +/- (square of b2 – 4ac) / 2a

X = - 28 +/- (square of 784 + 1548) / -9.8

X = - 28 +/- (square of 2332) / -9.8

X = -28 – 5

X = -33

- #6

berkeman

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You should write two separate equations -- one for each object with its initial conditions.

Can you simplify how to work on this? I'm just not getting it lol.

X =?

X_{o}= 79 M

Δt = ?

V_{o}= 28 m/s

A = -9.8 m/s^{2}

X (Δt) = X_{o}+ V_{o}Δt + .5a (Δt)^{2}

X (Δt) =79 m + 28/ms Δt + .5(-9.8 m/s^{2}) (Δt)^{2}

X (Δt) =79 m + 28/ms Δt - 4.9 m/s^{2}(Δt)^{2}

79 m (Δt) = 28 m/s Δt - 4.9 m/s^{2}(Δt)^{ 2}

Ax^{2}+ Bx + Cx = 0

-4.9 + 28 + 79 =0

X = -b +/- (square of b2 – 4ac) / 2a

X = - 28 +/- (square of 784 + 1548) / -9.8

X = - 28 +/- (square of 2332) / -9.8

X = -28 – 5

X = -33

1) A ball is dropped from a height of 79 m.

2) Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s.

Try starting with those two separate equations, and solve them simultaneously for the time t when the two x values are equal...

- #7

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Can you give me the two equation you stated?

Because my teacher told us that the initial height of an object is not important so we never used any equation to work on it.

And for the second equation, I'm not sure am I suppose to find the time, V

- #8

berkeman

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No, sorry. We cannot do your work for you -- that's against the PF rules (see the Rules link at the top of the page).

Because my teacher told us that the initial height of an object is not important so we never used any equation to work on it.

And for the second equation, I'm not sure am I suppose to find the time, V_{f}, or X.

The two equations you need to write are for the two objects. You apply this equation:

x(Δt) = Xo + VoΔt + .5a(Δt)2

to each object, based on the initial conditions you are given in the question. So like this (setting initial time t=0):

1) A ball is dropped from a height of 79 m.

x(t) = Xo + Vot + .5a(t)2

2) Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s.

x(t) = Xo + Vot + .5a(t)2

Fill out each of those two equations with the initial conditions you are given for each object (initial height, initial velocity [with correct sign]). Once you have those two equations, solve them simultaneously for the time t when the two x(t) values are equal

- #9

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x(Δt) = Xo + VoΔt + .5a(Δt)2

I never use this equation before so I have couple of question.

Why you add Δt to X?

What is Xo? is it the initial height?

Position of elevating object

x =?

Δt =?

Xo = ?

Vo = 28 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vot + .5a (Δt) ^2

X (Δt) = Xo + 28 m/s (Δt) + .5(-9.8 m/s ^2) (Δt) ^2

X (Δt) = Xo + 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

Did I plug in the value correctly? I feel like the information plugged in is too vague to deduce the next step.

- #10

berkeman

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The notation x(t) means the value of x as a function of time. If the time in your problem starts at t=0, then the delta-t just is the time t.x(Δt) = Xo + VoΔt + .5a(Δt)2

I never use this equation before so I have couple of question.

Why you add Δt to X?

What is Xo? is it the initial height?

Yes, in your notation using X for height, Xo is the initial height. What is it for object #1? What is it for object #2?

It would be better if you could keep the two objects' equations separate at first. Can you fill in the initial conditions for each in the format below?Position of elevating object

x =?

Δt =?

Xo = ?

Vo = 28 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vot + .5a (Δt) ^2

X (Δt) = Xo + 28 m/s (Δt) + .5(-9.8 m/s ^2) (Δt) ^2

X (Δt) = Xo + 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

Did I plug in the value correctly? I feel like the information plugged in is too vague to deduce the next step.

1) A ball is dropped from a height of 79 m.

x(t) = Xo + Vot + .5a(t)2

2) Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s.

x(t) = Xo + Vot + .5a(t)2

- #11

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Position of falling object

x =?

Δt =?

Xo = 79 m

Vo = 0 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vo Δt + .5a (Δt) ^2

X (Δt) = 79 m + .5 (-9.8) m/s ^2 (Δt) ^2

Position of elevating object

x =? (I don't think we know the height for the elevating object)

Δt =?

Xo = (Not sure is the initial height 0 m or yet to define)

Vo = 28 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vot + .5a (Δt) ^2

X (Δt) = Xo + 28 m/s (Δt) + .5(-9.8 m/s 2) (Δt) ^2

Is this the correct format for both equation?

- #12

berkeman

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This is correct for the object that is dropped from 79m up.Position of falling object

x =?

Δt =?

Xo = 79 m

Vo = 0 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vo Δt + .5a (Δt) ^2

X (Δt) = 79 m + .5 (-9.8) m/s ^2 (Δt) ^2

I don't know what this means...Position of elevating object

x =? (I don't think we know the height for the elevating object)

When you are on the ground, you are at 0m for initial vertical position. Set Xo to that.Δt =?

Xo = (Not sure is theinitial height 0 mor yet to define)

Vo = 28 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vot + .5a (Δt) ^2

X (Δt) = Xo + 28 m/s (Δt) + .5(-9.8 m/s 2) (Δt) ^2

Looks pretty good. Now set the two x(t) equal, because when that happens, the two objects will be passing each other. Solve away! You will likely get a quadratic equation that you have to solve, but just use your normal quadratic formula for that.Is this the correct format for both equation?

I have to bail for an hour or two, but will check back in later to see what you come up with.

- #13

berkeman

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BTW, I like the way you are carrying units along in your equations. That's an important trick that helps you to keep your equations balanced, and helps to catch mistakes where terms get dropped in calculations sometimes.

- #14

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Position of falling object

x =?

Δt =?

Xo = 79 m

Vo = 0 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vo Δt + .5a (Δt) ^2

X (Δt) = 79 m – 4.9 m/s ^2 (Δt) ^2

-79 m (t) = -4.9 m/s ^2 (Δt) ^2

16.1 s (t) = Δt ^2

16.1 s = Δt

Position of elevating object

x =?

Δt =?

Xo = 0 m

Vo = 28 m/s

A = -9.8 m/s ^ 2

X (Δt) = 0m + Vot + .5a (Δt) ^2

X (Δt) = 28 m/s (Δt) + .5(-9.8 m/s ^2) (Δt) ^2

X (Δt) = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

-28 m/s = - 4.9 m/s ^2

5.7 s

This is what I have after listening to your instruction, but I don't know what to do next.

- #15

berkeman

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No, you just solved for when both objects hit the ground (you set x=0 it looks like).

x =?

Δt =?

Xo = 79 m

Vo = 0 m/s

A = -9.8 m/s ^ 2

X (Δt) = Xo + Vo Δt + .5a (Δt) ^2

X (Δt) = 79 m – 4.9 m/s ^2 (Δt) ^2

-79 m (t) = -4.9 m/s ^2 (Δt) ^2

16.1 s (t) = Δt ^2

16.1 s = Δt

Position of elevating object

x =?

Δt =?

Xo = 0 m

Vo = 28 m/s

A = -9.8 m/s ^ 2

X (Δt) = 0m + Vot + .5a (Δt) ^2

X (Δt) = 28 m/s (Δt) + .5(-9.8 m/s ^2) (Δt) ^2

X (Δt) = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

-28 m/s = - 4.9 m/s ^2

5.7 s

This is what I have after listening to your instruction, but I don't know what to do next.

Take these two equations of yours:

X (Δt) = 79 m – 4.9 m/s ^2 (Δt) ^2

X (Δt) = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

and set them equal to each other. Do you see how that will be true at the time the two objects pass each other?

Once you set them equal to each other, solve for your Δt variable.

(okay, now I'm really leaving)

- #16

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Okay Thank you so much for your time..I will try to solve it and post again later.Good night

- #17

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I been trying for a while, and I don't know where to start off. I use quadratic formula in both equation, but the answer doesn't seem to make sense.

X (Δt) = 79 m – 4.9 m/s ^2 (Δt) ^2

With the use of quadratic equation, I acquire an answer of 8.1.

X (Δt) = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

With the use of quadratic equation, I acquire an answer of 2.8

I'm not sure how to find T, so not sure if I am even doing the right thing.

- #18

berkeman

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It looks like you are trying to solve the equations individually, instead of setting them equal (because the two heights will be equal when the objects pass each other). So you should take your two equations:

X (Δt) = 79 m – 4.9 m/s ^2 (Δt) ^2

With the use of quadratic equation, I acquire an answer of 8.1.

X (Δt) = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

With the use of quadratic equation, I acquire an answer of 2.8

I'm not sure how to find T, so not sure if I am even doing the right thing.

X (Δt) = 79 m – 4.9 m/s ^2 (Δt) ^2

X (Δt) = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

and set them equal to each other:

79 m – 4.9 m/s ^2 (Δt) ^2 = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

Now solve for your variable Δt...

- #19

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Okay.

Basically, I cancel out everything in this equation:

79 m – 4.9 m/s ^2 (Δt) ^2 = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2,

and got this:

79m = 28 m/s (Δt)

2.8 s = Δt

Then I replace 2.8 for t in both of the equation and check if my time is correct:

X = 79 m – 4.9 m/s ^2 (Δt)^2

X = 79 m – 4.9 m/s ^2 (2.8 s)^2

X = 79 m - 4.9 m/s ^2 (7.84s^2)

X = 79 m - 38.416 m

X = 40.6 m

X = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

X = 28 m/s (2.8s) - 4.9 m/s ^2 (2.8 s)^2

X = 78.4 m - 38.416 m

X = 40 m

Is this correct?

- #20

berkeman

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You rounded some of your answers. If you eliminate the rounding, I believe the answers would agree.

Basically, I cancel out everything in this equation:

79 m – 4.9 m/s ^2 (Δt) ^2 = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2,

and got this:

79m = 28 m/s (Δt)

2.8 s = Δt

Then I replace 2.8 for t in both of the equation and check if my time is correct:

X = 79 m – 4.9 m/s ^2 (Δt)^2

X = 79 m – 4.9 m/s ^2 (2.8 s)^2

X = 79 m - 4.9 m/s ^2 (7.84s^2)

X = 79 m - 38.416 m

X = 40.6 m

X = 28 m/s (Δt) – 4.9 m/s ^2 (Δt) ^2

X = 28 m/s (2.8s) - 4.9 m/s ^2 (2.8 s)^2

X = 78.4 m - 38.416 m

X = 40 m

Is this correct?

Good work!

- #21

HallsofIvy

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I'm still pondering the concept of two objects

- #22

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Opps lol!I'm still pondering the concept of two objectsfallingin opposite directions!

Thank you so much for your help Berkeman..Appreciate all the time you help me

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