- anonymous

Please, help? An Olympic diver springs off a high dive that is 3m above the surface of the water. When she lands in the water she is traveling at a speed of 8.9 m/s at an angle of 75.0° with respect to the horizontal. What was her take off speed and direction?

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- anonymous

@hartnn ?

- anonymous

- anonymous

picture!

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## More answers

- anonymous

|dw:1382313081201:dw|

- anonymous

First off, what do you know about her velocity in the x direction for the entire trajectory?

- anonymous

It's constant?

- anonymous

yup yup. So now all you need to find is her y velocity at 3m above the water, and you can figure out the direction and magnitude of initial takeoff , yah? How might one find that?

- anonymous

Ummm... I'm not sure

- anonymous

You have kinematics or energistics as your two options. Pick your poison.

- anonymous

or both

- anonymous

just not neither

- anonymous

Kinematics is what we're learning

- anonymous

that's probably the better option then :) What do you know about the diver's y velocity at the top of the trajectory?

- anonymous

8.9sin75° ?

- anonymous

That's right at the bottom, when she hits the water. What about right here?|dw:1382313886798:dw|

- anonymous

Ohh! 0

- anonymous

!!! :D

- anonymous

exactly

- anonymous

Did it all fall into place and the big picture come into focus? Or do wanna keep working it together?

- anonymous

I'm still lost tbh

- anonymous

none worries. Can you come up with an equation of motion that'll give you this height?|dw:1382314187399:dw|
One that only uses velocity, distance, and acceleration?

- anonymous

I'm not sure without t

- anonymous

There's 1 of em is all. It's
\[ v^2 = v_o^2 + 2ad\]
v = final velocity
vo = initial velocity
a = acc.
d = displacement

- anonymous

Hmmm I haven't even seen that yet!

- anonymous

That's the one I always forget, and always forget how to derive, too :P You get it (I'm neglecting a few terms, sorry) from
\[ d = \left( \frac{v_o + v}{2} \right)t \]
\[ t = \left( \frac{2}{v_o + v} \right)d \]
then plug that into
\[v = v_o + at\]
It's mighty handy. Anyways... Once you get height H, what can you do?

- anonymous

I could use this equation to find the velocity? But how would I find h?

- anonymous

you know the velocity though!
\[v = v_y= 8.9 sin \theta\]
at the bottom, and vo = 0 at the top.

- anonymous

Oh okay.

- anonymous

So below the diving board (or would it be below the surface of the water) the \(\ \large v_y = 8.9sin\theta \text{ ? }\)

- anonymous

exactly at the surface of the water Vy = 8.9sin75. We're trying to see, starting from rest, how far would the diver have to fall to be going that fast.

- anonymous

Does that make sense? I tend to make things more convoluted than they need to be :/

- anonymous

Yes, I think so. Where would I go from here?

- anonymous

with that, we have the last bit of info.
|dw:1382315519387:dw|
When we have H, we us that same equation to find out the magnitude of the velocity height "little h," which is in line with the diving board. Then using that value you can find the unknown initial angle φ (phi) using tanφ = y/x .
|dw:1382315666658:dw|
The magnitude of the initial velocity is then
\[ |\textbf v| = \sqrt{x^2+y^2}\]
Follow?

- anonymous

The answer takes the form, "Initial velocity was ...m/s at .... degrees above the horizontal."

- anonymous

Okay, just one moment...Im trying to understand this all...!

- anonymous

~~ The first paragraph should say "... find out the velocity ~~ at~~ height 'little h'..."

- anonymous

Okay that part of the velocity being same in direction but opposite in magnitude seems to make sense.... I'm a bit confused as to how to find x and y, though

- anonymous

you have x already. It's stayed the same throughout the trajectory. You get y by using that 4th equation of motion again.
\[ v_{y@h}^2 = 2ah\]
Then
\[\tan \phi = \frac{y}{x} = \frac{v_{y@h}}{8.9\cos7.5}. \]

- anonymous

oh, y and x components of the triangle. Not x and y as in displacement. Sorry :P

- anonymous

meant Vy and Vx

- anonymous

`~~
\[ tan \phi = \frac{v_{y@h}}{8.9 \cos75} \]

- anonymous

Hey, I gotta run. Sorry. I'll be on later :) good luck and good physicsing! Sciencing with friends is fun.

- anonymous

Ok! Thanks for the help so far! :)

- anonymous

change of plan. back. how's it goin?

- anonymous

I'm really stuck actually @AllTehMaffs!

- anonymous

I'm not sure which triangle x and y correspond to!

- anonymous

My bad, I kept throwing around variables willy nilly.
|dw:1382318816931:dw|
the triangle x and y - which I should have said Vx and Vy - correspond to is the triangle where V is the hypotenuse.
they're the x and y components of the initial velocity.
\[ v_ix = v_i \cos \phi \]
\[ v_iy = v_i \sin \phi \]
We just happen to know them already, and are calculating Vi based on them.
We know the initial velocity in the x direction is the same as the final x velocity down by the water
\[ v_ix = 8.9\sin75 \]
and the y velocity is what we found at little h - the velocity 3 meters above the water.
\[ v_iy = v_{y@h} = \sqrt{2ah}\]

- anonymous

This is starting to make some sense! What I'm confused about is how do I know what a and h are? Would h just be -9.8? Wait no 9.8? As for h... I'm not sure.

- anonymous

h = H - 3 meters
|dw:1382320100943:dw|
, and a is -9.81m/s^2. I'm playing kind of loose with the signs ;P .
When we find H:
\[ v_f^2 = 2aH\]
we're trying to see how big H is. We know H is a positive number, so we fix the units to be positive. This is a sloppy convention on my part. In that equation what we're really saying is
|dw:1382319969777:dw|
\[ v_f^2 = 2(-9.81)(-H)\]
That's how the units technically work.
Same for little h
|dw:1382320225339:dw|
We're measuring a distance below where we start "measuring."
...

- anonymous

Ahhh Okay

- anonymous

:)

- anonymous

So y is equal to \(\ \Large \sqrt{2 \times -9.8 \times h \text{ ?}} \)
\[\ \Large \text{What would h be then? I know it's H-3, but what's H? } \]

- anonymous

- anonymous

Wow, Thank You so much for all your help @AllTehMaffs! I finally understood how to find h, by first finding H using v^2f=2aH. I found h to be 1.037 (H=4.037), and I finally got an answer, which I think makes sense!!!!
I got initial velocity was 5.06 m/s at 62.9 degrees above the horizontal.
Thanks again @AllTehMaffs! :) :)

- anonymous

@Study23
Sooo close! All of the concepts are there, just make sure that you realize in the equation
\[ v_f^2=2aH\] (which, again, I wrote while using variables willy nilly)
**the only velocity that contributes to H, the maximum height the diver gets to, is the Y velocity!** so it should be
\[v_{f}^2 = v_o^2 + 2ad \]
\[v_{fy}^2 = \cancel{v_{oy}^2} + 2a(H) \]
\[v_{fy}^2 = 2a(H) \]
So H is
\[ H =\frac{v_{fy}^2}{2a}=\frac{{(8.9m/s\sin75)}^2}{(2)(-9.81m/s^2)}=-3.77m\]
This equation says, "Starting from rest, a body must travel -3.77m - it must fall 3.77m meters below an initial location y=0 to attain a velocity of (8.9 sin 75)m/s given that acceleration is -9.81 m/s^2.
A very wordy confusing say of looking at it!... which we can get around by saying everything is a magnitude. Then the equation reads as "Starting from rest, a body under acceleration 9.81 m/s^2 has a velocity of (8.9 sin 75)m/s in the direction of acceleration after it has displaced 3.77m in the direction of acceleration" Then all of the positive and negative information is retrieved from defining your coordinate system on the picture.
.
We can now say that the max height above the diving board , little h, is
\[ h = H - 3 = .77m\]
With that we're looking for the velocity of the diver after displacing .77 m
\[ v_{y@h}=\sqrt{2ah} = \sqrt{(2)(9.81m/s^2)(.77m)} = 3.76 m/s\]
"Starting from rest, after a body being accelerated at 9.81m/s^2 displaces .77m in the direction of the acceleration, it will be traveling a velocity of 3.76m/s in that direction."
Therefor, starting at the top of the trajectory and falling .77m (going .77m in the -y direction), its velocity will be 3.76 m/s in the negative y direction.
THIS means that at the other side of the parabola, its velocity will be equal in magnitude and opposite in direction! It's initial velocity in the y direction is 3.76 m/s in the positive y direction. Now we have v_xi and v_iy
\[ v_{xi} = v_{xf} = 8.9 \cos 75 m/s\]
\[v_{yi} = v_{y@h} = 3.76 m/s\]
Using these values to find our angle phi
\[ tan \phi = \frac{v_{yi}}{v_{xi}} = \frac{3.76m/s}{8.9 \cos 75 m/s}=1.59 \]
\[ \phi = arctan 1.56 = 57.34º\]
Now to find the magnitude of v_i, which is the square root of the sum of the squares of its components.
\[ |\textbf v_i| = \sqrt{|v_{xi}|^2+|v_{yi}|^2} = \sqrt {(8.9\cos75m/s)^2+(3.76 m/s)^2} = 4.4m/s \]
SO! ::breaths::
v_i = 4.4m/s @ 57.34º above the horizontal.
This makes physical sense. The angle entering the water should be steeper than the initial angle because of an increase in the magnitude of the v velocity due to gravity.

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