A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of 5.1 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.
Why did the criminal become an acrobat? Because he thought he could escape justice by jumping from rooftop to rooftop! Well, let's solve this puzzle for him.
To find the maximum value for D, we need to consider the time it takes for the criminal to reach the ground. Because we want him to land safely on the adjacent building, we need to calculate the time it takes for him to fall 2.0 m.
The equation for vertical displacement without air resistance is given by the formula:
δy = Viy * t + (1/2) * g * t^2,
Where δy is the vertical displacement, Viy is the initial vertical velocity, t is time, and g is the acceleration due to gravity.
Since the criminal is falling vertically, let's set δy = -2.0 m (negative because he's falling downwards), Viy = 0 (because he's not jumping upwards), and g = 9.8 m/s^2. Plugging these values in:
-2.0 = 0 * t + (1/2) * 9.8 * t^2,
Simplifying:
-2.0 = 4.9 * t^2.
Rearranging the equation:
t^2 = -2.0 / 4.9,
t^2 ≈ -0.408.
Since time cannot be negative, this means the criminal cannot reach the adjacent building safely. Sorry, Mr. Criminal! You might want to stick to a different escape plan.
To find the maximum value for D, we need to determine the horizontal distance the criminal can jump before hitting the ground.
Let's consider the horizontal and vertical motions separately.
Horizontal motion:
Since there is no horizontal force acting on the criminal, his horizontal velocity remains constant throughout the jump. This means that the time it takes for the criminal to reach the adjacent building is the same as the time it takes for him to fall vertically by 2.0 meters.
Vertical motion:
The criminal is free-falling vertically due to the force of gravity. We can use the equation of motion:
Δy = v0y * t + 1/2 * g * t^2
Given:
Δy = -2.0 m (negative because the criminal has fallen)
v0y = 0 m/s (the initial vertical velocity is zero)
g = 9.8 m/s^2 (acceleration due to gravity at the surface of the Earth)
Plugging in these values:
-2.0 = 0 * t + 1/2 * 9.8 * t^2
Simplifying the equation:
-2.0 = 4.9 * t^2
Solving for t:
t^2 = -2.0 / 4.9
t^2 = -0.408
Since time cannot be negative, this is not a possible solution. So, there is no possible time for the criminal to reach the adjacent building before hitting the ground.
Hence, the maximum value for D is 0 meters. The criminal cannot jump to the adjacent building without falling to the ground.
To find the maximum value for D, we need to determine the furthest distance the criminal can jump horizontally.
Since air resistance is negligible, there are no horizontal forces acting on the criminal once they leave the roof. Therefore, the horizontal component of the criminal's velocity remains constant throughout the jump.
We can use the equation of motion to find the horizontal distance traveled (D) by the criminal. The equation is:
D = Vx * t
Where:
- D represents the horizontal distance traveled
- Vx is the horizontal component of velocity
- t is the time of flight
To find the time of flight, we need to determine how long the criminal will remain in the air.
Since there is no vertical force acting on the criminal horizontally, we can use the equation for vertical motion under constant acceleration to find the time of flight. The equation is:
h = (1/2) * g * t^2
Where:
- h represents the vertical displacement (2.0 m)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time of flight
Rearranging the equation, we get:
t = sqrt((2 * h) / g)
Substituting the given values, we find:
t = sqrt((2 * 2.0) / 9.8) = 0.637 s (rounded to 3 decimal places)
Now, we can substitute the horizontal component of velocity (5.1 m/s) and the time of flight (0.637 s) into the equation for horizontal distance:
D = Vx * t = 5.1 m/s * 0.637 s = 3.25 m
Therefore, the maximum value for D is 3.25 meters.
h = Vo*t + 0.5a*t^2 = 2 m.
0 + 4.9t^2 = 2,
t^2 = 0.408,
Tf = 0.639 s. = Fall time.
Xo = 5.1 m/s = Initial Hor. velocity
D = Xo * Tf = 5.1m/s * 0.639s = 3.26 m.