What was the elapsed time of the flight stone?

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Homework Help Overview

The problem involves a stone thrown vertically upward from a tower, requiring the calculation of the total elapsed time of its flight until it strikes the ground. The context is kinematics, specifically dealing with projectile motion under gravity.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss various methods to calculate the time of flight, including the use of kinematic equations and the quadratic formula. Some question the initial setup and assumptions made about the height and motion of the stone.

Discussion Status

There are multiple interpretations of the problem and differing approaches to the calculations. Some participants have provided guidance on how to set up the equations, while others express confusion over the results obtained. The discussion is ongoing, with attempts to clarify the correct method and resolve discrepancies in the calculated times.

Contextual Notes

Participants note constraints such as the requirement to use specific kinematic equations and the implications of the stone's initial height. There is also mention of homework rules that may limit certain approaches.

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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!
 

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