How to solve a velocity problem as an initial condition problem

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A rock is thrown upward from a 30.0m tall building with an initial speed of 17.0m/s, and air resistance is negligible. The equations of motion are applied, with the acceleration due to gravity set at -9.81m/s². The initial position must be correctly defined as either 30m or 0m to avoid inconsistencies in calculations. The correct time for the rock to hit the ground is determined to be approximately 4.76 seconds, leading to a final velocity of -29.6m/s. Properly setting initial conditions is crucial for solving the problem accurately.
Wolvenmoon
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Homework Statement



A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 30.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.

Homework Equations



Acceleration of gravity = 9.81m/s^2

The normal way:
v^2 = u^2 + 2as
v=u+a*t
http://answers.yahoo.com/question/index?qid=20120204155622AAIs1sA

How I want to do it:
dv/dt = a(t)
ds/dt = v(t)

The Attempt at a Solution


a(t) = -9.8 m/s^2
v(t) = -9.8t m/s + C(m/s)
17 = -9.8*0 + C, C=17
v(t) = -9.8t(m/s) + 17(m/s)
s(t) = -4.9t^2 + 17t + C
30 = -4.9*0 + 17*0 + C
C = 30
s(t) = -4.9t^2+17t+30
-30 = -4.9t^2+17t+30

t=5.64s for part B.
For part A, this would make it -9.8*5.64 + 17 = -38.272.


My answers are wrong. Using the equations listed at Yahoo answers:
v=(17^2+2*9.81*30)^(1/2)=29.6m/s and 29.62=-17+9.81*t, t=4.76s

I'm certain there's a way to solve this as initial value problems. I'd rather get my calculus on than try to memorize these equations!
 
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Wolvenmoon said:
s(t) = -4.9t^2+17t+30
-30 = -4.9t^2+17t+30

Why -30 on the left hand side?
 
I believe a bad guess. The initial position is y=0m (the level of the roof), the ball ends up at y=-30.

That probably should be a zero.
 
You need to decide. Either s(0) = 30, which fixes one value for the integration constant. Or s(0) = 0, which fixes some other value. Then you should have s(T) = 0 or S(T) = -30, respectively. But s(0) = 30, which you chose originally, is not compatible with with s(T) = -30 (because that would mean the street is 60 meters below the 30 meter tall building, which is impossible).
 
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Thank you! That does provide the correct answer in either case (t=4.76, then that can be substituted into v(t) = -29.6m/s!)

I'm glad to know it was something simple.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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