Problem solving: ball drops & ball shot up - they meet where?

[Tyrannosaurus_]

Homework Statement

Ball 1 falls from rest from 67 m up. Directly 67 m below, ball 2 has 20 m/s [up]. No air. Where do they meet?

Homework Equations

non-uniform motion, so key kinematics equations.

The Attempt at a Solution

d1 + d2 = 67 m <--- is this okay to do?
v1i *t + 1/2 a *t² + V2i *t + 1/2 a *t² = 67 m
0 *t + g/2 *t² + V2i *t + g/2 *t² = 67 m
2(g/2 *t²) + V2i *t = 67 m
g *t²) + 20 *t - 67 = 0

Then, quadratic formula... this definitely isn't working. Any suggestions would be greatly appreciated.

 

[scottdave]

Ball#1 is accelerating in the same direction as it's motion. Second ball acceleration is opposite direction of initial motion.
The initial velocity 20 is up, while gravity points down, so they need to be opposite signs.

 

[PeroK]

Ball#1 is accelerating in the same direction as it's motion. Second ball acceleration is opposite direction of initial motion.
The initial velocity 20 is up, while gravity points down, so they need to be opposite signs.

... which also means that relative to each other the balls are not accelerating.

@Tyrannosaurus_ does that mean you can solve this problem without considering acceleration?

 

[Tyrannosaurus_]

Thanks for the help everyone.
I neglected the negative sign for g by mistake. When I use -9.8 in the quadratic formula, I get a negative radical/under the root. So, I suspect the dilemma is in the setup.

I have considered acceleration - it's g as there is no air resistance. All equations are for non-uniform motion. Both objects accelerate in the same direction, i.e., down. Despite the equivalent acceleration for both, it is the initial velocity of the ball leaving the ground that allows them to close their separation.

In the quadratic formula, I am substituting:
a = -9.8
b = +20
c = -67

This result provides imaginary roots. So, I suspect it's in the set-up, but I'm stuck.

 

[PeroK]

Thanks for the help everyone.
I neglected the negative sign for g by mistake. When I use -9.8 in the quadratic formula, I get a negative radical/under the root. So, I suspect the dilemma is in the setup.

I have considered acceleration - it's g as there is no air resistance. All equations are for non-uniform motion. Both objects accelerate in the same direction, i.e., down. Despite the equivalent acceleration for both, it is the initial velocity of the ball leaving the ground that allows them to close their separation.

In the quadratic formula, I am substituting:
a = -9.8
b = +20
c = -67

This result provides imaginary roots. So, I suspect it's in the set-up, but I'm stuck.

If negative is down, then ##-67m## is below the ground.

 

[Tyrannosaurus_]

Thanks for the help PeroK.

I understand where you're coming from, but do you see why I have a negative value for c. It's from d1 + d2 = 67 m. It's really the purpose for my OP, how can I understand that initial statement in a better way so that I can find the place where the two balls collide.?

 

[PeroK]

Thanks for the help PeroK.

I understand where you're coming from, but do you see why I have a negative value for c. It's from d1 + d2 = 67 m. It's really the purpose for my OP, how can I understand that initial statement in a better way so that I can find the place where the two balls collide.?

The first thing you could do is calculate the position of each ball separately as a function of time.

 

[Tyrannosaurus_]

Thanks for the suggestion PeroK! That will work for me.

I'm really curious why I can't find an algebraic solution though. Should my original equation look like:
-d1 + d2 = 67 m?

I suggest a negative displacement for ball 1 as it is moving down, and ball 2 is moving up. I've set up as +, down as -.
But, that still leaves me wondering if the total displacement from both balls would be + 67 m, or -67 m. I'm going to plot the functions, but an algebraic understanding would really help me going into midterms, etc.

Thanks PF community!

 

[PeroK]

Thanks for the suggestion PeroK! That will work for me.

I'm really curious why I can't find an algebraic solution though. Should my original equation look like:
-d1 + d2 = 67 m?

I suggest a negative displacement for ball 1 as it is moving down, and ball 2 is moving up. I've set up as +, down as -.
But, that still leaves me wondering if the total displacement from both balls would be + 67 m, or -67 m. I'm going to plot the functions, but an algebraic understanding would really help me going into midterms, etc.

Thanks PF community!

You mixed up two approaches. The two displacements do not add up to either ##+67m## or ##-67m## as one displacement is positive and the other is negative.

You could, however, have considered the distance traveled by each ball. The two distances should add up to ##67m##. Although you would need to be careful that the ball thrown up did not reach its highest point and then fall back down before the dropped ball collided with it.

 

[Sammit Dhar]

d1+d2=67m
Let's call the distance between the top of the tower(I have assumed it was dropped from the top of a tower) and the place where they meet "x"

For Ball1,
x=1/2 gt^2

For Ball2,
67-x=20t-(1/2 gt^2)

Add these two, and you get
67-x+x=20t-(1/2 gt^2)+(1/2 gt^2)

67=20t

You get t, and then you can work out x from Ball1's or Ball2's equation.

Hope I was helpful.

 

[Tyrannosaurus_]

d1+d2=67m
Let's call the distance between the top of the tower(I have assumed it was dropped from the top of a tower) and the place where they meet "x"

For Ball1,
x=1/2 gt^2

For Ball2,
67-x=20t-(1/2 gt^2)

Add these two, and you get
67-x+x=20t-(1/2 gt^2)+(1/2 gt^2)

67=20t

You get t, and then you can work out x from Ball1's or Ball2's equation.

Hope I was helpful.

Wow! Thanks for the helpful reply! Very helpful!

I solved it a little while ago in a similar way to you:

For ball 1,

H(t) = 67m - 1/2gt^2

For ball 2,
H(t) = 20t - 1/2 gt^2

Then set, H(t) = H(t) and solved for t. Found the same result as you, i.e., t = 67/20 s

Thanks for the help Sammit! (I think there are too many people on here posting for the sake of posting. Of all the posts, only yours offers anything of value. Thank you!)

 

[neilparker62]

... which also means that relative to each other the balls are not accelerating.

@Tyrannosaurus_ does that mean you can solve this problem without considering acceleration?

Useful insight here - means you can write down directly: 20t = 67