Physics Question
A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of 5.0 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.
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ANSWERS: 1
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Remember your kinematics equations: at constant velocity (5.0 m/s, in this case), x = vx t, where x is the horizontal distance, vx is the horizontal velocity, and t is the time. Also, for a vertical drop, h = 1/2 g t2, where h is height, g is gravity (9.8 m/s2), and t2 is time squared. Just fill in the numbers and plug away. x = D, vx = 5.0 m/s, h = 2.0 m, so x = vx t becomes D = 5 m/s t and h = 1/2 g t2 becomes 2 m = 1/2 9.8 m/s2 t2. Substitute t from the first equation into the second one and: 2 m = 1/2 9.8 m/s2 (D/(5 m/s))^2. Solve algebraically for D... 2 m = 4.9 m/s2 D2 / 25 m2/s2, so D2 = 100/9.8 m2. D = 3.2 m. Hopefully his bones are strong.
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