Question

Please Help! Will give medal and fan for best answer!


Imagine that you are standing at the edge of a cliff. You throw one ball straight upward and another ball straight downward at the same time and initial speed. Neglecting the effects of air resistance, compare the speeds of the balls when they hit the ground below the cliff?

Answer 1

So you know that the initial velocities are given as
\[v_{i1} = +v\]
\\ v_{i2} = -v\]
and the distance at which you're evaluating the final velocities is given by
\[\Delta y = -d\]
that is, a distance d below where the rock started.
You can compare their final velocities given the fourth equation of motion
\[v_{f1}^2 = (v_{i1})^2 - 2ad
\\ \quad =(+v)^2 - 2ad
\\ \
\\ \
\\ v_{f2}^2 = (v_{f1})^2-2ad
\\ \
\\ \quad = (-v)^2 - 2ad\]

What does that give? ^_^

Answer 2

oops, sorry
the equation should be of the form
\[v_f^2=v_i^2 + 2(-a)(-d)
\\ \quad = v_i^2 + 2ad\]

Answer 3

That first bit is messed up, my apologies! The equations give
\[(v_{f1})^2 = (v_{i1})^2+2ad
\\ (v_{f1})^2 = (+v)^2+2ad
\\ \
\\ \
\\ (v_{f2})^2 = (v_{i2})^2+2ad
\\ (v_{f2})^2 = (-v)^2+2ad\]

Answer 4

(the perils of morning maths)

Answer 5

ok so..... im confused. Im really bad at math and really bad at Physics. so what am i supposed to do?

Answer 6

My terrible typesetting isn't helping things :P

So, we first define each velocity to be some value, v
\[v_{1i} = v
\\ v_{2i} = -v\]
If it helps, you can assign a completely arbitrary value, ie
\[v_{1i} = 10m/s
\\ v_{2i} = -10m/s\]
You just know that they're equal in magnitude and opposite in direction.

Then the fourth equation of motion in the y direction is is
\[v_f^2 = v_i^2 + 2a\Delta y\]
Where acceleration is gravity
\[a=-9.81m/s^2\]
and the distance traveled is the height of the cliff, with your starting point up top as the zero point
\[\Delta y =y_f-y_i=0-d= -d\]
again, you can arbitrarily assign a height if it helps, ie
\[\Delta y = -50m\]

Then all you need to do is compare the final velocity of the upwards object with the final velocity of the downwards object by plugging in the values for V1i and V2i, respectively.

Answer 7

ooooohhhhh I get it! (i think)! :0

Answer 8

So
\[v_{1f}^2 = v_{1i}^2 + 2a\Delta y
\\ = (+v)^2+2(-9.81m/s^2)(-d)
\\ =(10m/s)^2 + 2(-9.81m/s^2)(-50m) \leftarrow \text{these are the example values}\]
and
\[v_{2f}^2 = v_{2i}^2 + 2a\Delta y
\\ = (-v)^2+2(-9.81m/s^2)(-d)
\\ =(-10m/s)^2 + 2(-9.81m/s^2)(-50m) \leftarrow \text{these are the example values}\]

Answer 9

Then using those example values, what is the value for V1f? For V2f?

Answer 10

i have no idea. I'm Sorry you are doing a great job trying to explain this to me. But i dont get it. So am I just going to do this
=(10m/s)2+2(−9.81m/s2)(−50m)
and that will give me the answer to V1F?

Answer 11

No worries ^_^

Those equations give gives the final velocity squared of v1 and v2.

So
\[v_{1f}^2 = v_{1i}^2 + 2a\Delta y
\\ = (+v)^2+2(-9.81m/s^2)(-d)
\\ =(10m/s)^2 + 2(-9.81m/s^2)(-50m) \leftarrow \text{these are the example values}
\\ \
\\ v_{1f} = \sqrt{(10m/s)^2 + 2(-9.81m/s^2)(-50m)}
\\ = \sqrt{10^2 + 2(-9.81)(-50)} \ m/s\]

Similarly for v2
\[v_{2f}= \sqrt{(-10)^2+2(-9.81)(-50)} \ m/s\]

Have you seen that equation of motion before?
the
\[v_{f}^2 = v_{i}^2 + 2a\Delta y\]

?

Answer 12

yes im learning about t but need some more help understanding it. Its really confusing for me.

Answer 13

Do you want to derive it together real quick? It's only a few steps, and it might help in understanding ^_^

Answer 14

ok sure
thank you so much

Answer 15

Sorry, technical difficulties :P

So first, far a given object with a final and an initial velocity from a constant acceleration (in the y direction for here), the displacement (how far it moves, again in the y direction) during time t is given by
\[y_f = y_i + \left(\frac{v_i+v_f}{2}\right) t\]
does that one look familiar?

Answer 16

The average of the velocities multiplied by time, plus the initial position is equal to the final position.

Answer 17

(goodness, the website is not cooperating this morning :P )

Answer 18

hey you can totally solve this without any equations you know :P

Answer 19

hey do you want a non mathematical approach?!

Answer 20

Just by saying what the velocity of the upwards particle is on its way back down at the initial position?

Answer 21

Yeah, can you snag this @mashy, I've probably convoluted this enough for one morning :P

Answer 22

bingo! ;-).. without having to get our hands dirty in brute maths.. no offence.. i know u love it :P

Answer 23

curry girl.. are u with us? dozed off?

Answer 24

This site is bottle necked!! geez!

Answer 25

^_^

And yeah, the site is definitely having trouble thinking this mornin'!

Answer 26

Bah, I gotta run anyway. Thanks !

(but since it's already typed out, here's the end of the derivation!)
From there, we can rearrange that equation to solve for time
\[y_f-y_i = \left(\frac{v_i+v_f}{2}\right) t\]
\[\Delta y = \left(\frac{v_i+v_f}{2}\right) t\]
\[ \Delta y \left(\frac{2}{v_i+v_f}\right) = t\]
\[t = \Delta y \left(\frac{2}{v_i+v_f}\right)\]

Next, we also know the equation for final velocity given an acceleration

\[v_f = v_i + at\]
The final velocity is equal to the initial velocity plus the amount of time the object is being accelerated. We can plug in that value of t into this equation!
\[v_f = v_i + at\]
\[v_f = v_i + a\left( \Delta y \left(\frac{2}{v_i+v_f}\right) \right)\]
\[(v_f)(v_f+v_i) = v_i(v_f+v_i) + 2a\Delta y\]
\[v_f^2 + \cancel{v_fv_i} = v_i^2 + \cancel{v_fv_i} + 2a\Delta y\]
\[ v_f^2 = v_i^2 + 2a\Delta y\]
Voila!