Calculating Time Until Balls Reach Same Height

AI Thread Summary
To determine when two balls reach the same height, one dropped from 79 m and the other thrown upward at 28 m/s, the equations of motion for both must be established. The first ball's position is given by x(t) = 79 - 4.9t², while the second ball's position is x(t) = 28t - 4.9t². Setting these equations equal allows for solving the quadratic equation for time t. The calculations indicate that the two balls meet after approximately 2.21 seconds. Understanding the initial conditions and correctly applying the kinematic equations is crucial for solving such problems.
louis676
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1. A ball is dropped from a height of 79 m. Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s. How much time passes until the balls are at the same height.

I'm not really sure what equation to use or how to approach.
 
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louis676 said:
1. A ball is dropped from a height of 79 m. Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s. How much time passes until the balls are at the same height.

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

Welcome to the PF.

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...
 


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 physics 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 physics 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 = VoΔt + .5a(Δt)2

Δx = ?
Vo = 0 m/s
a = -9.8 m/s 2
Δt = ?

Δx = VoΔt + .5a(Δt)2
Δx = (0m/s)(Δt) + .5(-9.8m/s2)(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 = VoΔt + .5a(Δt)2

Δx = ?
Vo = 28 m/s
a = -9.8 m/s 2
Δt = ?

Δx = VoΔt + .5a(Δt)2
Δx = 28 m/s (Δt) + .5(-9.8 m/s2) (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.
 


louis676 said:
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 physics 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 physics 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 = VoΔt + .5a(Δt)2

Δx = ?
Vo = 0 m/s
a = -9.8 m/s 2
Δt = ?

Δx = VoΔt + .5a(Δt)2
Δx = (0m/s)(Δt) + .5(-9.8m/s2)(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 = tVertical position of object throwing up
Δx = VoΔt + .5a(Δt)2

Δx = ?
Vo = 28 m/s
a = -9.8 m/s 2
Δt = ?

Δx = VoΔt + .5a(Δt)2
Δx = 28 m/s (Δt) + .5(-9.8 m/s2) (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.


Usually we label vertical position with "y", and horizontal position with "x", but that doesn't really matter in this problem because you are dealing only with one axis (up-down).

You need to include the initial "x" position in your basic equation. That is what makes this problem much easier to do.

So instead of just this:

Δx = VoΔt + .5a(Δt)2

You need to include the initial position:

x(Δt) = Xo + VoΔt + .5a(Δt)2

Try setting up the two equations using this full version of the position versus time kinematic equation...
 


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 =?
Xo = 79 M
Δt = ?
Vo = 28 m/s
A = -9.8 m/s 2

X (Δt) = Xo + VoΔt + .5a (Δt) 2
X (Δt) =79 m + 28/ms Δt + .5(-9.8 m/s2) (Δt) 2
X (Δt) =79 m + 28/ms Δt - 4.9 m/s2 (Δt) 2
79 m (Δt) = 28 m/s Δt - 4.9 m/s2 (Δt) 2
Ax2 + 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
 


louis676 said:
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 =?
Xo = 79 M
Δt = ?
Vo = 28 m/s
A = -9.8 m/s 2

X (Δt) = Xo + VoΔt + .5a (Δt) 2
X (Δt) =79 m + 28/ms Δt + .5(-9.8 m/s2) (Δt) 2
X (Δt) =79 m + 28/ms Δt - 4.9 m/s2 (Δt) 2
79 m (Δt) = 28 m/s Δt - 4.9 m/s2 (Δt) 2
Ax2 + 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

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

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...
 


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, Vf, or X.
 


louis676 said:
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, Vf, or X.

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).

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)22) Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s.
x(t) = Xo + Vot + .5a(t)2Fill 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
 


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


louis676 said:
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?
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.

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?

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.

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?

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


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


louis676 said:
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
This is correct for the object that is dropped from 79m up.

Position of elevating object

x =? (I don't think we know the height for the elevating object)
I don't know what this means...
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
When you are on the ground, you are at 0m for initial vertical position. Set Xo to that.

Is this the correct format for both equation?

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.

I have to bail for an hour or two, but will check back in later to see what you come up with.
 
  • #13


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


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


louis676 said:
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.

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

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) :smile:
 
  • #16


Okay Thank you so much for your time..I will try to solve it and post again later.Good night
 
  • #17


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


louis676 said:
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.

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

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


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


louis676 said:
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?

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

Good work!
 
  • #21


I'm still pondering the concept of two objects falling in opposite directions!
 
  • #22


HallsofIvy said:
I'm still pondering the concept of two objects falling in opposite directions!

Opps lol!

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

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