Question

A security guard is chasing a burglar across a rooftop, both running at \(4.20 \mathrm{~m} / \mathrm{s} .\) Before the burglar reaches the edge of the roof, he has to decide whether or not to try jumping to the roof of the next building, which is \(5.50 \mathrm{~m}\) away and \(4.00 \mathrm{~m}\) lower. If he decides to jump horizontally to get away from the guard, can he make it? Explain your answer.

Short answer

Answer: No, the burglar cannot make the jump successfully, as they can only cover a horizontal distance of approximately 3.79 m while falling the 4.00 m vertical distance.

Step-by-step solution

Calculate the time it takes for the burglar to fall 4.00 meters

We can use the free-fall equation to find the falling time. The equation is: \(y = \frac{1}{2}gt^2\) where: \(y\) is the vertical distance (\(4.00 \mathrm{~m}\)), \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m} / \mathrm{s}^2\)), and \(t\) is the time. We need to solve for t: \(t = \sqrt{\frac{2y}{g}}\) Plug in the known values, \(y = 4.00 \mathrm{~m}\) and \(g = 9.81 \mathrm{~m} / \mathrm{s}^2\): \(t = \sqrt{\frac{2(4)}{9.81}}\)

Calculate the time

Calculate the time it takes to fall the 4 meters using the values inputting in the equation. \(t = \sqrt{\frac{8}{9.81}}\) \(t \approx 0.903 \mathrm{~seconds}\)

Calculate the horizontal distance covered during the falling time

We can use the time (\(t = 0.903 \mathrm{~seconds}\)) and the horizontal speed (\(4.20 \mathrm{~m} / \mathrm{s}\)) to find out the horizontal distance the burglar can jump: \(d = vt\) where: \(d\) is the horizontal distance, \(v\) is the horizontal speed, and \(t\) is the time. Plug in the known values, \(v = 4.20 \mathrm{~m} / \mathrm{s}\) and \(t = 0.903 \mathrm{~seconds}\): \(d = 4.20 * 0.903\)

Calculate the horizontal distance

Calculate the horizontal distance the burglar would be able to cover by using the horizontal speed and time. \(d = 4.20 * 0.903\) \(d \approx 3.79 \mathrm{~m}\)

Compare the horizontal distance covered with the distance between the buildings

Now we'll compare the calculated horizontal distance the burglar can jump (\(3.79 \mathrm{~m}\)) and the distance between the buildings (\(5.50 \mathrm{~m}\)). If the horizontal distance covered is greater than or equal to the distance between the buildings, then the burglar can make the jump. Since \(3.79 \mathrm{~m} < 5.50 \mathrm{~m}\), the burglar cannot make the jump successfully.

Key concepts

Free-Fall Motion

When we talk about free-fall motion, we refer to the motion of an object under the sole influence of gravity. This is a key concept in physics, particularly in mechanics, and it's essential to understand it when solving problems related to objects falling from a height.

Any object in free fall near the Earth's surface accelerates downward at a constant rate, which we call the acceleration due to gravity, denoted by the symbol \( g \) and has a value of approximately \( 9.81 \text{ m/s}^2 \). The intriguing part about free-fall is that all objects, regardless of their mass, fall at the same rate if we neglect air resistance.

To determine the time it takes for an object to fall a certain distance, you can use the equation \( y = \frac{1}{2}gt^2 \), where \( y \) is the vertical distance fallen, and \( t \) is the time in seconds. By rearranging this equation to solve for \( t \), we can calculate the fall time, which is crucial for understanding the context within scenarios like the burglar attempting to jump between buildings.

Kinematic Equations

Kinematic equations allow us to describe the motion of an object in terms of its initial velocity, final velocity, acceleration, time, and displacement. Understanding these equations is crucial when analyzing any motion scenario, such as a burglar trying a daring leap between rooftops.

The kinematic equations are derived from the basic principles of motion and are especially useful for objects moving with constant acceleration, like those in free-fall. One of these equations is:\[ v = u + at \], where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the constant acceleration, and \( t \) is time. Another important equation is \( s = ut + \frac{1}{2}at^2 \), wherein \( s \) represents the displacement.

In our burglar's scenario, the kinematic equations allow us to calculate the distance he can cover horizontally during the fall. By understanding these equations, we can apply them to both horizontal motion (at constant speed, since there's no horizontal acceleration) and vertical free-fall motion (with acceleration \( g \) due to gravity).

Projectile Motion

Projectile motion involves objects moving in two dimensions under the influence of gravity. In these scenarios, an object's motion occurs in both the horizontal and the vertical planes simultaneously, with gravity affecting only the vertical component of the motion. This concept is key when assessing whether our burglar can make the jump from one rooftop to another.

When we delve into projectile motion, we typically assume the initial vertical velocity is zero (unless stated otherwise), as is the case in a horizontal jump. The horizontal component of velocity remains constant throughout the flight, as there's no horizontal acceleration (neglecting air resistance).

Combining the constant horizontal velocity with the time it takes to fall the vertical distance, we can compute the horizontal range using the equation \( d = vt \), with \( d \) being the horizontal distance, \( v \) the horizontal speed, and \( t \) the time. By dissecting the motion into its horizontal and vertical components, we can solve complex projectile motion problems, like figuring out the safe landing for our fleeing burglar.