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?(adsbygoogle = window.adsbygoogle || []).push({});

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

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