- #1
ccgrad05x2
- 11
- 0
A ball is thrown upward from the top of a 25.4-m-tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 32.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?
This problem will use the kinematics equations.
Here is what I have so far:
First, I will want to determine the time that the ball will spend in the air. Then, I can divide the distance by the time and that will give me the average speed of the person.
First, I must account for the time the ball will spend going up after being thrown. I calculated this to be:
V = Vinitial + at
0 = 12.0 + (-9.80xt)
t- 1.22449
I can then multiply this number by 2 to account for the ball coming back down to the "top of the building" So far, total time is 2.44898.
Now I must account for the time the ball is falling down the 25.4 m.
Using this formula: displacement = vinitialt+ .5at^2
-25.4 = 12.0t + .5(-9.8)t^2
0 = .4.9t2 +12.0t + 25.4
I solved for this, and found the time to be 3.80965 seconds.
That would give me a total time of 3.80965 + 2.44898 = 6.25863
Then, Avg Speed = Distance/Time = 32.0/6.25863 = 5.11924 m/s
This answer is not correct though, and I am having trouble figuring out what I am missing.
Thanks for the help
This problem will use the kinematics equations.
Here is what I have so far:
First, I will want to determine the time that the ball will spend in the air. Then, I can divide the distance by the time and that will give me the average speed of the person.
First, I must account for the time the ball will spend going up after being thrown. I calculated this to be:
V = Vinitial + at
0 = 12.0 + (-9.80xt)
t- 1.22449
I can then multiply this number by 2 to account for the ball coming back down to the "top of the building" So far, total time is 2.44898.
Now I must account for the time the ball is falling down the 25.4 m.
Using this formula: displacement = vinitialt+ .5at^2
-25.4 = 12.0t + .5(-9.8)t^2
0 = .4.9t2 +12.0t + 25.4
I solved for this, and found the time to be 3.80965 seconds.
That would give me a total time of 3.80965 + 2.44898 = 6.25863
Then, Avg Speed = Distance/Time = 32.0/6.25863 = 5.11924 m/s
This answer is not correct though, and I am having trouble figuring out what I am missing.
Thanks for the help