What is the Maximum Horizontal Distance for a Rooftop Escape?

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SUMMARY

The maximum horizontal distance (D) a criminal can achieve when jumping off a rooftop at a speed of 5.7 m/s, with the adjacent building 2.0 m below, is calculated using the physics of projectile motion. The time of flight is determined by the formula t = √(2s/g), where s is the vertical distance (2 m) and g is the acceleration due to gravity (9.81 m/s²). The correct time of flight is approximately 0.447 seconds, leading to a maximum horizontal distance of approximately 2.6 meters. The initial calculation of 2.324 meters was incorrect due to a miscalculation in the time of flight.

PREREQUISITES
  • Understanding of basic kinematics and projectile motion
  • Familiarity with the equations of motion, specifically s = ut + 0.5at²
  • Knowledge of gravitational acceleration (g = 9.81 m/s²)
  • Ability to perform square root calculations
NEXT STEPS
  • Review the principles of projectile motion in physics
  • Practice solving problems involving horizontal and vertical motion separately
  • Learn how to derive time of flight for different projectile scenarios
  • Explore the effects of air resistance on projectile motion
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion dynamics.

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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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