A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of 5.6 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.
To find the maximum value for D, we need to determine the horizontal distance the criminal can jump before falling 2.0 m below the starting point.
Let's consider the horizontal and vertical motion separately.
For horizontal motion, the criminal is moving at a constant speed of 5.6 m/s. Therefore, the time it takes for the criminal to travel any distance horizontally will be the same.
For vertical motion, we have an initial vertical displacement of 0 m (since the criminal jumps horizontally), an initial vertical velocity of 0 m/s (since the criminal jumps horizontally), and a vertical displacement of -2.0 m (the height difference between the two rooftops). We need to find the time it takes for the criminal to reach this vertical displacement.
To find the time of flight, we can use the kinematic equation for vertical motion:
ωy = ωyo + ay * t,
where ωy is the final vertical velocity, ωyo is the initial vertical velocity, ay is the vertical acceleration (which is the acceleration due to gravity and is approximately -9.8 m/s^2), and t is the time of flight.
Since the criminal jumps horizontally, the initial vertical velocity, ωyo, is 0 m/s.
Using the equation:
0 = 0 + (-9.8) * t,
we can solve for t:
-9.8t = 0,
t = 0.
Hmm, it appears that the time of flight is 0 seconds. This means the criminal falls instantaneously. As such, the maximum value for D is 0 meters (assuming the buildings are right next to each other).
However, if there were no height difference between the rooftops or if the criminal had a vertical velocity component when jumping, the time of flight would be different, allowing for a non-zero maximum value for D.