⦁ A criminal is trying to escape across a rooftop and runs off the roof horizontally at a speed of 4.9 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 1.9 m below the jumping-off point. Find the maximum value for D.
h = 0.5g*t^2 = 1.9
4.9t^2 = 1.9
t^2 = 0.388
Tf = 0.623 s. = Fall time.
D = Xo*Tf = 4.9m/s * 0.623s = 3.05 m.
To find the maximum value for D, we need to determine the horizontal distance the criminal can jump before landing on the adjacent building.
We can solve this using the equations of motion. We'll assume that the initial height of the criminal is zero (at the same level as the adjacent building's roof) and the acceleration due to gravity is 9.8 m/s².
First, we need to calculate the time it takes for the criminal to reach the level of the adjacent building's roof when falling vertically. We can use the kinematic equation:
Δy = v0y * t + 0.5 * a * t²
Since Δy = -1.9 m (negative because the criminal is falling), v0y = 0 (initial vertical velocity), and a = -9.8 m/s² (acceleration due to gravity), we can rearrange the equation to solve for t:
0.5 * (-9.8) * t² = -1.9
Simplifying:
4.9 * t² = 1.9
t² = 1.9 / 4.9
t ≈ 0.383 s
Now that we have the time it takes for the criminal to reach the same height as the adjacent building's roof, we can calculate the horizontal distance traveled during that time. We'll use the equation:
Δx = v0x * t
Since the criminal runs horizontally with a velocity of 4.9 m/s, v0x = 4.9 m/s:
Δx = 4.9 * 0.383
Δx ≈ 1.8787 m
Therefore, the maximum value for D, the horizontal distance between the buildings, is approximately 1.8787 meters.