My physics instructor said that this is not intuitive but a correct assumption to make. I cant figure out why this must be. Assume the objects are released at exactly the same time. They meet at the halfway point. Why do they necessarily have the same velocity? And does this apply if they don't meet halfway (equal distance)?

Assume air resistance is negligible.

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    $\begingroup$ This sounds incorrect as stated. Are you sure you heard your teacher right? What exactly did they say? $\endgroup$
    – knzhou
    Aug 23 '16 at 21:31
  • $\begingroup$ @knzhou One of my peers was working a problem that said, "one object is dropped from height h and another is thrown upwards with initial velocity $v_0$. They meet at height h/2. What is the initial velocity of the thrown object in terms of h?" and he assumed that the two object's velocities at the meeting point are the same. I asked the instructor if this was a correct assumption to make and he said yes, but it's unintuitive as to why, and he couldn't explain it. The student got the right answer under the assumption of equal velocity. $\endgroup$
    – rb612
    Aug 23 '16 at 21:37
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    $\begingroup$ If you draw a velocity vs. time graph, you can see that the velocity of the dropped object at the time the two objects meet is equal to the initial velocity of the thrown object $v_0$. But it is not equal to the velocity of the thrown object at the meeting point. $\endgroup$
    – knzhou
    Aug 23 '16 at 22:05
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    $\begingroup$ Why don't you try to solve the problem by using the equations of motions rather than guessing? In this way you will find if the assumption is right. There is also the assumption that they meet in the air. Do you think this is right to assume? $\endgroup$
    – nasu
    Aug 23 '16 at 22:10

This is not true. Let's see this in full details:

If one body is being droped and the other is being thrown up, forgetting about air resistance, the only important force is the weight of the bodies.

So, for starters, lets consider one body of mass $m$. The weight is $\mathbf{F} = -mg\mathbf{\hat{y}}$. Now, applying Newton's second law, we get the equation of motion:

$$m\mathbf{a} = -mg\mathbf{\hat{y}} \Longrightarrow \mathbf{a} = -g\mathbf{\hat{y}}.$$

Integrating we get the velocity:

$$\mathbf{v}(t)=-gt \mathbf{\hat{y}}+\mathbf{v}_0,$$

where $\mathbf{v}_0$ is the initial velocity. Integrating once more we get the displacement:


where $\mathbf{r}_0$ is the initial position. So to apply this to the particular problem, we imagine the reference frame is so that $y = 0$ is the ground.

In that case, we have two bodies:

  • The first body has mass $m_1$ and is droped from height $h$ with no initial velocity. Thus $\mathbf{r}_0 = h\mathbf{\hat{y}}$ and $\mathbf{v}_0=0$.

  • The second body has mass $m_2$ and is thrown straight up from the ground with initial velocity $v_0$ so that $\mathbf{v}_0= v_0 \mathbf{\hat{y}}$.

The question is, therefore, do they meet if the motion for both starts at $t=0$? Meeting means, of course, $\mathbf{r}_1(t)=\mathbf{r}_2(t)$.

So the equation is:


This is just the simple equation:


Thus, they meet when $t = h/v_0$. Now, do they have the same velocity when they meet? To answer that we compute $\mathbf{v}_1$ and $\mathbf{v}_2$ at this time:



Thus we have equality just when


which means $h = 0$. Because of that, this cannot happen.


The final velocity of the dropped ball is the same as the initial velocity of the thrown ball because

  • they experience the same acceleration
  • they travel the same vertical distance
  • they take the same time to travel the same distance.

without all of these three conditions the velocities are (most likely) different.

Going through equations to prove this...

For the dropped ball

  • initial velocity, $u_d$, is equal to zero
  • final velocity, $v_d$, is unknown
  • time of meeting is $t$
  • distance travelled is $h/2$
  • acceleration is $g$

so (using equation $v^2=u^2+2as$)

$$ v_d ^2 = 2gh/2 = gh \\ v_d = \sqrt {gh}$$

(and using equation $t = (v - u) / a$, which can be rearranged to $v=u+at$)

$$ t = {\sqrt {gh} \over g} = \sqrt {h \over g}$$

now for the thrown ball

  • $u_t$, initial velocity is unknown.
  • $v_t$, final velocity is unknown.
  • $a = -g$ - the acceleration is $g$ again, but now the acceleration is reducing the velocity so that it needs a minus sign.
  • $t$, the time is the same as the time above so $t=\sqrt{h\over g}$
  • $s$, distance is again $h/2$

rearranging the equation $s=ut+{1 \over 2}at^2$ we get $u = {s - {1 \over 2}at^2 \over t}$


$$u_t= {h/2 - {1 \over 2}(-g){h \over g} \over \sqrt {h \over g}} \\ = {h/2 + {1 \over 2}{h} \over \sqrt {h \over g}} \\ = \sqrt {h g}$$

Thus, the equations tell us that $u_t = v_d$ - always, for the three conditions listed at the top.

Finally, it is easy to show that $v_t$ the final velocity of the thrown ball must be zero. I am not going to prove it mathematically, but it is not difficult to do from here.

Now thinking about it the dropped ball starts at zero and ends up with some velocity... whereas the thrown ball ends up with zero velocity because the distance, time and acceleration are the same (except the acceleration is negative for the thrown ball).

If you plot a distance time graph for the two objects you will find that they have similar (similar here = identical) parabolic shapes(, but sort of reflected in the y-axis. )

Hope this is helpful...

  • $\begingroup$ This is awesome! Thank you so much!! So I did the math to verify that vf = 0, and did $\frac{d}{dt}\sqrt{gh}t-4.9t^2 = 0$ so $\sqrt{gh}-gt=0$ and got that $t = \sqrt{h/g}$ so it all works out! Thanks again. $\endgroup$
    – rb612
    Aug 24 '16 at 1:25

Start by drawing a very simple diagram of the situation and then see where these two equations lead you. Apply them to each object separately. Then compare results:

$$V_f^2 - V_o^2 = 2ax$$

$$x = \frac{1}{2}at^2 + V_ot + x_o$$

Another hint: The quadratic equation might be applicable in your proof of truth or falsehood.

  • 1
    $\begingroup$ So the final velocity of the thrown object is $V_f^2 = 2(-9.8)\frac{L}{2} + V_0^2$ while the final velocity of the dropped object is $V_f^2 = 2(-9.8)\frac{-L}{2}$ so I'm not quite sure how to prove they'd be unequal or equal. $\endgroup$
    – rb612
    Aug 23 '16 at 21:49
  • $\begingroup$ @rb612 See my edit. $\endgroup$ Aug 23 '16 at 21:52

According to your comment, the problem is :

One object is dropped from height $h$. At the same instant another is thrown upwards from the ground with initial velocity $v_0$. They meet at height $\frac12h$. What is the initial velocity $v_0$ of the thrown object in terms of h?

When they meet the 2 objects have travelled the same distance $\frac12h$ in the same time. Both accelerate downwards at $g$, so they gain/lose the same velocity $v_1$ in the same time. When they meet, the velocity of the dropped object has increased to $0+v_1$, while the velocity of the object thrown upwards decreases to $v_0-v_1$.

The distances and times travelled up to this point are equal, so the average velocities are also equal (because acceleration is constant). Therefore
hence $v_0=v_1$.

When they meet the velocities are $0+v_1=v_0$ for the dropped object and $v_0-v_1=0$ for the thrown object. They are not equal at the meeting point. They have switched over.

This can be visualised on the velocity-time graph. The slope of the two graphs is the same in magnitude (opposite in sign). The areas under the 2 graphs do not become equal (the condition for the meeting point) until the objects have switched velocities.

enter image description here

Since the dropped object falls through height $\frac12 h$ and reaches velocity $v_0$ then
$v_0^2=2g(\frac12 h)=gh$.

Perhaps your classmate misunderstood the velocity-time graph and thought that the objects meet when the graphs intersect. This is true on a displacement-time graph but not on a velocity-time graph. The velocities are equal at this instant, but the objects do not meet here.

At this point both objects have velocity $\frac12 v_0$. So if the dropped object has fallen through height $\frac12 h$ and reached speed $\frac12 v_0$ then
$(\frac12 v_0)^2=2g(\frac12 h)=gh$

This is not the same answer as above. So I don't see how your classmate got the same answer using his assumption.


Well, if we consider that they move at a constant velocity, they wouldn't meet halfway after a certain time if they hadn't the same velocity...

instant 0 :


instant t* :

In the case objects 1 and 2 have the same velocity :


Example of a case with V1>V2 :


Edit : nvm

  • $\begingroup$ But one of the masses is drop (from height $h$), which implies an initial velocity of zero. $\endgroup$
    – fibonatic
    Aug 23 '16 at 22:01

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