# What was the elapsed time of the flight stone?

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 ammount 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

HallsofIvy
Homework Helper
"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.

HallsofIvy