Help with a Physics Problem
M.M. Dalkilic
August 5, 2007
My suggestion in working physics problems is writing the formula first without the values–this helps you
remember the formula and cuts down on errors.
• Given: An object is thrown upward with velocity v and reaches a height of h. At that moment, a
second object is thrown upward with velocity v.
• Question: Describe where the two objects pass each other.
Answer
As everyone always suggests, draw a picture of the problem if you can. Fig. 1. shows my depiction. The
problem seems to be that we need to find two distances whose sum is h:
1. the object dropping from the top of h that has traveled x distance (I simply decided to call this distance
x–you could’ve called it anything)
2. the object rising from the bottom that has traveled h − x.
Figure 1: Drawing the problem to better understand it. The distance traveled is h and the initial velocity is v.
Once the object reaches h, a second object is thrown upward–where do they cross? It’s clear they pass somewhere–I
decided at x. Read throught the text before trying to make sense of the mathematics on the right.
1
You can verify that their sum x + (h − x) = h; so, we need to find this point.
Here’s my thinking: I know the top object has fallen x units and from that can calculate how long it
took to travel this far, i.e. t. I then use this time to figure out how far the object thrown from the ground
has traveled during that time. I’ll need to find the initial velocity v0 too, but that shouldn’t be too difficult
since I know how far it traveled...h. I’ll break this up into two parts. (1) finding the time x traveled; (2)
using this time to figure about h − x.
Before I begin, I want to list the formulua that I used:
d = vi + 1/2at2
(1)
vf2 = vi2 + 2ad
(2)
d = vi t + 1/2at2
(3)
I’ll explain now all the variables.
d is distance
vi is initial velocity
vf is final velocity
a is acceleration due to gravity
t is time
One last thing to remember is that there is direction–so if you’re going up, acceleration due to gravity will
be negative, and if you’re going down, it’ll be positive. Now finally the solution. We need to find the time
the object fell, the starting velocity from the bottom, then bring them together.
• Find the time the top object traveled x.
d = vi + 1/2at2
2x
= vi + t2
a
2x
= 0 + t2
ar
2x
=t
a
2
• We now find the original starting velocity
vf2 = vi2 − 2ad
√
2ah = vi
(4)
(5)
A couple of points to observe. The acceleration due to gravity is working against the original velocity,
so we subtract it in (4). The final velocity is 0 when the object reaches its maximum height (5).
• We can now bring everything together.
d = vi t + 1/2at2
r
√
2x
2x 2
h − x = 2ah
− 1/2a(
)
a
a
r
22 ahx
2x
h−x=
− 1/2a
a
a
√
h − x = 2 hx − x
r
h2 = 4hx
h = 4x
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