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.
d = Vo*t + 0.5g*t^2 = 2 m.
0 + 4.9t^2 = 2
t^2 = 0.408
Tf = 0.639 s. = Fall time.
D = 5.0m/s * 0.639s = 3.19 m.
To find the maximum value for D, we need to determine the distance the criminal can jump horizontally before landing on the adjacent building.
Let's analyze the projectile motion of the criminal.
First, let's find the time it takes for the criminal to land on the adjacent building. We can apply the kinematic equation:
h = v_iy * t + (1/2) * g * t^2
Since the roof of the adjacent building is 2.0 m below the jumping-off point, the initial vertical velocity v_iy is 0 m/s (since the criminal jumps horizontally). The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Thus, the equation simplifies to:
h = (1/2) * g * t^2
Plugging in the values:
2.0 = (1/2) * 9.8 * t^2
Simplifying the equation:
t^2 = 2.0 / (0.5 * 9.8)
t^2 = 0.408
t ≈ 0.639 seconds
Now that we know the time it takes for the criminal to land, we can determine the horizontal distance traveled by using the equation:
D = v_ix * t
The initial horizontal velocity, v_ix, is given as 5.0 m/s.
Plugging in the values:
D = 5.0 * 0.639
D ≈ 3.195 meters
Therefore, the maximum value for D is approximately 3.195 meters.