Difficult Free Fall/Kinematic Problem

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The problem involves two stones: the first is dropped from a building, and the second is thrown downwards 2.4 seconds later with an initial speed of 33.0 m/s, landing simultaneously with the first. To solve for the time it takes the first stone to reach the ground, the relationship between the times of both stones must be established, with the second stone's time being 2.4 seconds less than the first. The distances fallen by both stones are equal, allowing the use of free fall equations to relate their motions. The acceleration due to gravity is consistently 9.8 m/s² for both stones, and by setting up the equations correctly, the problem can be solved for the height of the building and the speeds just before impact. The discussion emphasizes the importance of correctly relating the variables and understanding the effects of gravity on both stones.
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Homework Statement



A first stone is dropped from the roof of a building. 2.40 s after that, a second stone is thrown straight down with an initial speed of 33.0 m/s, and it is observed that the two stones land at the same time.

(a) How long did it take the first stone to reach the ground?
_____________ s
(b) How high is the building?
_____________ m
(c) What are the speeds of the two stones just before they hit the ground?
_____________ m/s (first stone)
_____________ m/s (second stone)

Homework Equations



Free Fall Equations:
Vf = g*t
d= .5*g*t^2

and
Kinematic Equations
U1L6a1.gif


The Attempt at a Solution



If I knew any variables, I might be able to use the Free Fall Equations, but no variables are given for the first stone. The second stone only tells me Vi and a little bit of information about t..but not enough to get me started really. I'm truly lost on this one. I'm sorry my attempt at a solution seems pethedic! :frown:
 
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first, you have to use all the values they give you. If you let t_1 represent the time for the 1st stone then t_2, the time for the 2nd stone, is equal to t_1 - 2.4sec. Can you figure out what the values for the d's should be? or would it be easier to relate the two d's like how I related the 2 t's. Also, can you figure out what the value for the acceleration should be?
 
The distances for both stones are the same since they're being dropped from the same height..I tried setting the equations equal to each other but it said my answer was wrong.

Would the acceleration for the first stone be 9.8 m/s2? For both? I'm not sure my answer keeps on being wrong.
 
Write d=vt+1/2 at^2 for both stones.
You know that d is the same
You know v for both stones
You know a. Remember that although one stone was thrown, as soon as it left your hand it isn't being accelarated by you anymore, only by gravity - exactly the same as droping it.

You will get an equation with 2 unknown times, but you know how the times are related T1=T2+2.4 so you have only one unknown, you can solve for T and then use the same equation to find d.
 
Thanks! It took me awhile, (made many silly little mistakes), but I finally got it! =)
 
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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