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.
See previous post.
To find the maximum value for D, we need to determine the minimum velocity the criminal needs to have when jumping off the roof. Let's break down the problem and use kinematic equations to solve it.
1. First, let's consider the horizontal motion of the criminal. Since there is no horizontal force acting on the criminal once they jump off the roof, their horizontal velocity will remain constant throughout the motion. Therefore, the velocity (Vx) remains 5.0 m/s.
2. Now, let's consider the vertical motion of the criminal. The criminal jumps off the roof with an initial vertical velocity (Vy). Since the criminal wants to land on the roof of the adjacent building, the vertical displacement (Δy) should be -2.0 m (negative because the criminal is landing lower).
3. We can use the kinematic equation for vertical motion to relate the initial vertical velocity, displacement, and the acceleration due to gravity.
Δy = Vy * t + (1/2) * g * t^2
where Δy is the vertical displacement (-2.0 m), Vy is the initial vertical velocity (unknown), t is the time of flight, and g is the acceleration due to gravity (-9.8 m/s^2).
4. Since the criminal doesn't have any vertical motion after landing on the adjacent building, the time of flight (t) can be determined using the formula:
Δy = Vy * t + (1/2) * g * t^2
Plugging in the values:
-2.0 = Vy * t - 4.9 * t^2
5. To maximize the horizontal distance D, the criminal should have the minimum possible time of flight. This occurs when the criminal reaches the maximum height (at the top of the trajectory) just as they reach the edge of the roof. Therefore, the time of flight (t) is equal to half of the total time spent in the air.
t = T/2
where T is the total time of flight.
6. The total time of flight (T) can be calculated using the vertical motion formula:
Vy = 0 - g * T/2
Rearranging the equation:
T = -2Vy / g
7. Substituting the value of T/2 into the equation from step 5, we get:
t = (-2Vy / g) / 2
t = -Vy / g
8. Substituting the expression for t from step 7 into the equation from step 4, we can solve for Vy:
-2.0 = Vy * (-Vy / g) - 4.9 * (-Vy / g)^2
9. Simplifying the equation from step 8, we have a quadratic equation in terms of Vy:
-2.0 = -Vy^2 / g + 4.9 * Vy^2 / g^2
10. Rearranging the equation from step 9:
-2.0 * g^2 = -Vy^2 * g + 4.9 * Vy^2
Simplifying further:
2.0 * g^2 = 4.9 * Vy^2 + g * Vy^2
10.4 * Vy^2 = 2.0 * g^2
Vy^2 = (2.0 * g^2) / 10.4
Vy = sqrt((2.0 * g^2) / 10.4)
11. Plugging in the value of g = 9.8 m/s^2 into the equation from step 10, we can calculate Vy:
Vy = sqrt((2.0 * 9.8^2) / 10.4) ≈ 4.41 m/s
12. Since the horizontal velocity (Vx) remains constant at 5.0 m/s, we can use the Pythagorean theorem to find the resultant velocity (V) of the criminal:
V = sqrt(Vx^2 + Vy^2)
V = sqrt((5.0)^2 + (4.41)^2)
V ≈ 6.49 m/s
13. Finally, we can calculate the maximum value for D using the formula for horizontal motion:
D = V * t
D = 6.49 m/s * (-Vy / g)
Plugging in the values:
D = 6.49 m/s * (-4.41 m/s) / 9.8 m/s^2
D ≈ 2.91 m
Therefore, the maximum value for D is approximately 2.91 meters.