What was the elapsed time of the flight stone?

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The discussion revolves around calculating the elapsed time of a stone thrown vertically upward from a tower. The initial approach involved using the equation of motion but led to incorrect results due to miscalculations. Participants emphasized the importance of incorporating the tower's height into the equation and solving for when the height equals zero. The correct method involves using the quadratic formula to find the positive solution for time. Ultimately, the consensus is that the elapsed time for the stone's flight is approximately 5.9 seconds.
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Here's the question, A stone is thrown vertically upward with a velocity of 26.0 m/s from the top of a tower having a height of 18.5m. On it's return it misses the tower and finally strikes the ground. What was the elapsed time of the flight stone?


I started if off with delta y= 26t+1/2(-9.91)t^2. From there I derived that forumula getting 26+0.81t=0. Solved for t, got 2.65s. I plugged t into the first formula and got 34.35. To get the other half of the time the stone was in the air I used 26t_2+1/2(-9.81)t_2^2. I solved for t_2 and added it with 34.45, but didn't end up with the correct answer. Where did I go wrong?
 
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From you calculations I can't see your "battle plan". here's mine.
1. find out how much time does it take the stone to stop. v=g*t1
2. it takes it the same amount of time to fall to the top of the tower.
3. find out how long does it take from here to hitting the ground. h=v*t2 + g*t22/2.
4. finally t=2*t1+t2
 
Your solution is almost right. First instead of using
"delta y= 26t+1/2(-9.91)t^2", you can simply add the known initial height (18.5 m) to say y= 18.5+ 26t+1/2(-9.91)t^2 where y is height above the ground.

Now, DON'T differentiate to get velocity: you don't need to find the highest point, that's not asked. Since y IS the height above the ground, the problem is simply asking "what is t when y= 0?" In other words, solve the equation 18.5+ 26t+1/2(-9.91)t^2= 0.
Since this is a quadratic, it will have two solutions. You only want the positive one.
 
I alright I tried what you said and ended up with 46.14 s. Is that correct?
 
The Mentors's answer is perfect. You dervived a relation when you are not allowed to and you did it wrong. By solving the second degree equation, as I and my friend did, the result is 5.9 s. Using the day to day numerical values for the heigt and the speed of the stone, this amount of time is really belivable.
 
I have no idea how you got "46.14 s" but it is certainly not correct. Plugging it into the equation,18.5+ 26t+1/2(-9.91)t^2 = 0,
we get 18.5+ 26(45.14)+1/2(-9.91)(46.14)^2= 18.5+ 1199.64- 10548.70=
-9330.66 NOT 0!

Use the quadratic formula!
 
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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