An Olympic diver on a 10.0-m platform makes a dive by leaving the platform with an initial speed of 1.50 m/s upward. Answer the following:
A) How fast is the diver going when she reaches the water?
B) How high above the pool does the diver reach at apex?

I think I know the answers, but I'm not sure and I don't have anyone to check against. Anyone here taken AP Physics?

Yes

I'm taking AP physics right now, but that problem is like the stuff I did last year in Honors physics.

Don't you have an equation to go by? H(T) = Something.

You have to take both? I didn't take Honors Physics.

a(t) = -9.8 m/s²

Integrating gives velocity and position without fail (v₀ = 1.5, x₀ = 10), and it's a little more math than I want to do right now.

x(t) = ∫∫a(t) dt² = -4.9t² + 1.5t + 10

and graph it.

.. just use kinematics. dtbk. stop giving bad tips.
for first part:
v0 = 1.5 m/s down (up and down are same magnitude)
vf = we're looking for that
change in x: 10
gravity = -9.8
BLARG you have answer.

A) 14.09 m/s
B) 10.115 m


DTBK that's the wrong approach because we're not looking for time. Use the chain rule :

a = dv/dt = (dv/dx)(dx/dt) = dv/dx*(v)
adx = vdv
∫adx = ∫vdv
ax = v^2/2-v_0^2/2
v^2 = v_0^2 + 2*a*x
plug in the values:
0 = 1.5^2 + 2*(-9.81)*x

x = 0.115m
add 10m to it, you get 10.115 m which is the answer to part B)

then

v^2 = v_0^2 + 2*a*x
v^2 = 0 + 2(-9.81)(-10.115)
v = 14.09 m/s

My way is definitely the fastest if you have a graphing calculator.. I'll race you

I was just saying the first thing that came into my head.

How would finding time which is totally irrelevant in this problem make it faster? You're just wasting time and doing unnecessary work.

v^2 = v_0^2 + 2*a*x is all you need. I was only proving how i derived my equation.


(and remember, you can't always let your graphing calculator do all the work, you have to show your work on paper )

This solution is the best.

Time is always what I instinctively solve first, because it usually shows a lot of information. Because with time on your side, your a rolling stone.

In Physics: Mechanics, we've been taught to write down our given variables and to state our unknowns. With those, we decide which kinematic equation to use. Saves time especially when we've only got an hour to take the test . However, when acceleration isn't constant and becomes a function of time, distance, or even velocity, that's where integration comes into play.

It's a trick question. The diver never makes it to the water because his initial speed isn't enough to move him out from over the board.

Also, MA, you're wrong. B is 10.1. There's only 3 significant figures, or to the tenths, depending on the way you solve it.

She can't, there's an elephant in the way

For a problem like this, you obviously assume that after the diver jumps off the platform, it explodes.
Physics is cool like that.

Haha so true

That's what I would do. I'd solve for t first though and then use that to get the final velocity.

Post has been edited 1 time(s), last time on Sep 10 2008, 10:30 pm by Mayor.

There was some fancy equation (but pretty simple), but you could probably do it via intuition and knowledge of what gravity is.

Or, as DTBK said, graph it. I loved Kinematics back when I was going over this two years ago, though. It was simple, easy to imagine and followed obvious rules. It was easy to see when you need it. Not like kilowatts and stuff. I already forgot all of that circuitry .
Lies. The problem never specified that there was no horizontal (x) velocity.

I'd use x = v_it + 1/2at^2
Solve for t and then use v_f = v_i + at to get the final velocity.

Or use v_f^2 = v_i^2 + 2ax setting v_f to 0 and solving for x to get the the distance traveled from the board to the max height. Then using the max height you can solve for v_f using v_f^2 = v_i^2 + 2ax with v_i set to 0. Correct me if I'm wrong.

Post has been edited 1 time(s), last time on Sep 10 2008, 10:56 pm by Mayor.

That's how I did it.