What is the Maximum Horizontal Distance for a Rooftop Escape?

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A criminal jumps off a rooftop horizontally at 5.7 m/s, aiming to land on an adjacent building 2.0 m below. The initial calculations for the time of flight used the formula s=ut + 0.5at^2, leading to a time of approximately 0.4077 seconds. The horizontal distance calculated was 2.324 m, but this was identified as incorrect. A suggestion was made to check the math, specifically the square root calculation in the time formula. Correcting the calculations is essential for determining the maximum horizontal distance D accurately.
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Homework Statement


A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of 5.7 m/s, hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is D, and the roof of the adjacent building is 2.0 m below the jumping-off point. Find the maximum value for D.


The Attempt at a Solution


s=ut + .5at^2, so t^2 = 2s/g = (2 x 2)/9.81, t=0.4077, then do .4077 x 5.7 = 2.324 m, but that answer is wrong. I have no idea what is going on here, please help.
 
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xupe33jrm said:

Homework Statement


A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of 5.7 m/s, hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is D, and the roof of the adjacent building is 2.0 m below the jumping-off point. Find the maximum value for D.


The Attempt at a Solution


s=ut + .5at^2, so t^2 = 2s/g = (2 x 2)/9.81, t=0.4077, then do .4077 x 5.7 = 2.324 m, but that answer is wrong. I have no idea what is going on here, please help.

Check your math, did you take the square root of (2 x 2)/9.81?
 


It was my math, thanks! I could not figure out what I was doing wrong!
 

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