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.

Answer 1

Answer:

D = 3.0 m

Explanation:

• Assuming no other forces than gravity acting upon the criminal, his horizontal velocity must keep constant at 5.0 m/s.
• So, the horizontal displacement, applying the definition of average velocity, is just:

        $\Delta x = v_{h} * t (1)$

• In the vertical direction, the criminal is in free fall, starting from rest.
• Since we know the dfference in height between both roofs, we can use the kinematic equation for vertical displacement, as follows:

        $\Delta h = \frac{1}{2} * g * t^{2} = 2.0 m$

• Solving for t, we get:

        $t =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*2.0m}{9.8m/s2} } = 0.6 s (2)$

• Replacing in (1), we get:

       $\Delta x = v_{h} * t = 5.0 m/s * 0.6 s = 3.0 m$

• So, if the horizontal distance between the two buildings is larger than 3.0 m, the criminal will fall to ground.