Is the ADA's Sample Physics Problem Incorrect?

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The discussion centers on a physics problem from the ADA regarding a ball thrown horizontally from a building. The original poster calculates the building's height as 62m, while the ADA states it is 31m. A participant clarifies that the horizontal velocity is constant, and the correct time of flight should be calculated to solve for height. The correct formula includes a factor of 0.5, which the original poster initially overlooked. The poster expresses relief upon realizing the mistake and appreciates the help received.
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Hi, I think the sample physics problem given by the ADA on their website is wrong. Can anyone confirm? It says: A ball is thrown horizontally at 20 m/s from the top of a building and strikes level ground 50 m from the building. How high is the building?

When I solve this, I get 62m, but the answer is 31m. I assume the discrepancy is the horizontal velocity. I'm assuming it is constant while maybe it should be an average? Can anyone help me confirm this? I thought I had a strong hold on kinematics and I'm worried I'm getting stuck on this problem.

Thank you, anyone!
 
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burapi said:
Hi, I think the sample physics problem given by the ADA on their website is wrong. Can anyone confirm? It says: A ball is thrown horizontally at 20 m/s from the top of a building and strikes level ground 50 m from the building. How high is the building?

When I solve this, I get 62m, but the answer is 31m. I assume the discrepancy is the horizontal velocity. I'm assuming it is constant while maybe it should be an average? Can anyone help me confirm this? I thought I had a strong hold on kinematics and I'm worried I'm getting stuck on this problem.

Thank you, anyone!

The answer on the website is correct, and the horizontal velocity is meant to be constant. It seems you do not use the correct equation.
 
The time of flight, assuming contant horizontal velocity, is:

t = 50m/(20 m/s) = 2.5s.

Using s = u*t + 0.5 * a * t^2 you get 31 m.

Getting 62m looks like you forgot the 0.5.
 
My guess is you used the vertical velocity at impact instead of the average vertical velocity...
 
Yes, thank you for pointing out I used the wrong equation. I did indeed leave out the 1/2. Ironically I am relieved and can move on!

You guys are so great :)
 
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