Do Balls Thrown Up and Dropped Down Meet Above the Halfway Point of a Building?

[UMich1344]

Homework Statement

A ball is dropped from rest from the top of a building and strikes the ground with a speed v_f. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is v₀ = v_f, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building, above the halfway point, or below the halfway point. Give your reasoning.

Homework Equations

Kinematics equations.

The Attempt at a Solution

Strictly conceptually, we know that when something is thrown up, its initial velocity is greater than its final velocity (at its peak). When something is dropped, its initial velocity is less than its final velocity (at the ground).

We can conclude that when an object travels with a greater velocity over a certain amount of time than a second object does, the larger the displacement of the first object will be.

When something is dropped, it speeds up as its displacement becomes larger. On the other hand, when something is thrown up, it slows down as its displacement becomes larger.

With this reasoning we can conclude that the balls will cross paths above the halfway point of the building because the ball that is thrown up will have a greater displacement than the ball that is dropped when they cross paths.

Is that correct?

I want to back up my answer with some math work too. However, I'm not sure how to go about it. I know I can refer to the height of the building as h and then solve for the point at which the two balls cross and compare my answer to h in order to see if the point is greater than 0.5h. I just don't know exactly how to set it up.

 

[Galadirith]

Hi UMich1344, yeh I think you logical conclusions are fine there so that's looks. So this does in fact again simply boil down to our lovely equations of motion (for constant acceleration :D).

So first we need to determine when our properties of each ball, so well call the balled dropped ball A and the ball thrown up ball B:

Ball A:
- u = 0 ms^-1
- s = s m
- t = t
- a = -9.8 ms^-2

now notice that we don't have a final velocity of actual values for s or t, that's because we want and equation of displacement at time t. (just as a note, actually these equations will not give relative displacement to the balls initial positions, but give and position vertically :D)

Now we should also note that the initial position of Ball A is at h if we consider that the bottom of the building is the origin of our coordinate system, if we call the height of the building h, so we end up with equations:

$$s = ut + \frac{1}{2}at^2 + s_i$$

note that we have s_i at the end, they are not part of the standard equations of motion you will most likely learn, but is simply the initial position, as in most cases the inital position is 0.

$$x_A(t) = -4.9t^2 + h$$

Now for B:
- u = v_f ms^-1
- s = s m
- t = t
- a = -9.8 ms^-2

so we get:

$$x_B(t) = {v_f}t -4.9t^2$$

now we want to find the point where they meet each other so we can say that:

$$x_A(t) = x_B(t)$$

that will yield a value for t (have a go at finding it :D). now obviously that is no good, we want to find a value for the position at which they meet in terms of h.

For that we simply substitute the value back into one of our equations for the position of a ball, it doesn't matter which one as both will give the same result (you can check that for yourself if you like)

Now at this point you may find it a little strange, as if you have done it all correctly, you will fin that you value of t was in terms of h and v_f and therefore you value of the position will also be in terms of them, but well don't want v_f so how do we get rid of it.

Well again we go back to our equations of motion. v_f is equal to the speed when ball A has fallen a distance h under the influence of gravity or:

$$v^2 = u^2 + 2as$$
$$v_f^2 = 2(-9.8)(-h)$$
$$v_f^2 = 19.6h$$

note that this time h is given as being negative, as this time we are not dealing with an coordinate system relative to a common origin, we are simply finding a final velocity with a displacement relative to the object, and the displacement is down so negative if this case.
so see if you can now use this final equation to get a value for the position of the balls in terms of h :D have fun

 

[thomate1]

Your reasoning is correct. To further support your answer with math, you can use the kinematic equations to solve for the time and displacement of each ball.

For the ball that is dropped, we can use the equation:

vf^2 = v0^2 + 2ad

where vf is the final velocity (which is equal to the initial velocity of the second ball, v0), v0 is the initial velocity (which is 0 since the ball is dropped from rest), a is the acceleration due to gravity (9.8 m/s^2), and d is the displacement (which we will solve for).

Rearranging the equation, we get:

d = (vf^2 - v0^2) / 2a

Substituting in the values, we get:

d = (vf^2 - 0^2) / 2(9.8)

d = vf^2 / 19.6

For the ball that is thrown upward, we can use the equation:

vf = v0 + at

where vf is the final velocity (which is equal to the initial velocity of the first ball, vf), v0 is the initial velocity (which is also vf), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time (which we will solve for).

Rearranging the equation, we get:

t = (vf - v0) / a

Substituting in the values, we get:

t = (vf - vf) / -9.8

t = 0

This means that the two balls will cross paths at t = 0, which is the same time as when the first ball hits the ground.

To solve for the displacement of the second ball at t = 0, we can use the equation:

d = v0t + (1/2)at^2

Substituting in the values, we get:

d = vf(0) + (1/2)(-9.8)(0)^2

d = 0

This means that the second ball will be at the height of 0 at t = 0, which is the same height as when the first ball is dropped.

Therefore, the two balls will cross paths at the same height, which is above the halfway point of the building.