Question

A car drives straight off the edge of a cliff that is \(60.0 \mathrm{~m}\) high. The police at the scene of the accident note that the point of impact is \(150 . \mathrm{m}\) from the base of the cliff. How fast was the car traveling when it went over the cliff?

Short answer

Answer: The car was traveling at approximately 42.87 m/s when it went over the cliff.

Step-by-step solution

Determine the time

Since the vertical motion is uniformly accelerated, we can use the following equation to find the time it takes for the car to fall to the ground: \(h = \frac{1}{2}gt^2\) where \(h\) is the height of the cliff, \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)), and \(t\) is the time it takes for the car to fall. Solving for time: \(t = \sqrt{\frac{2h}{g}}\) Substituting the given values: \(t = \sqrt{\frac{2(60)}{9.81}}\)

Calculate time using the given values

Calculate the time based on the height of the cliff and the acceleration due to gravity: \(t = \sqrt{\frac{120}{9.81}} \approx 3.499 \mathrm{~s}\)

Determine the car's initial horizontal speed

Since the horizontal motion is uniform, we can use the following equation to find the initial horizontal speed of the car: \(x = vt\) where \(x\) is the horizontal distance and \(v\) is the initial horizontal speed of the car. Solving for the initial horizontal speed: \(v = \frac{x}{t}\) Substituting the given values: \(v = \frac{150}{3.499}\)

Calculate the speed of the car

Calculate the initial speed of the car using the horizontal distance and the time it took for the car to hit the ground: \(v = \frac{150}{3.499} \approx 42.87 \mathrm{~m/s}\) The car was traveling at approximately \(42.87 \mathrm{~m/s}\) when it went over the cliff.

Key concepts

Kinematics

Kinematics deals with the motion of objects without considering the forces that cause the motion. In this problem, we're interested in understanding how the car moves from the top of the cliff to the point where it hits the ground. Kinematics allows us to break down motion into two main components: displacement and time.
By analyzing the vertical and horizontal motion separately, we use specific kinematic equations that relate these components. For the vertical motion, the equation used is based on the concept of free fall, while for horizontal motion, uniform motion equations come into play. This separation simplifies complex motion into simple, solvable equations.

• Vertical motion considers the change in position from the cliff to the ground.
• Horizontal motion involves the distance from the base of the cliff to the point of impact on the ground.

Understanding kinematics provides a systematic way to solve problems involving moving objects like the car in this exercise.

Uniformly Accelerated Motion

Uniformly accelerated motion describes a constant acceleration over time. In our scenario, the vertically falling car experiences uniformly accelerated motion due to the gravitational pull of the Earth. The acceleration due to gravity is approximately \(9.81 \text{ m/s}^2\).
This type of motion is characterized by specific equations, one of which is used in this exercise to determine the time it takes for the car to fall. The formula \( h = \frac{1}{2}gt^2\) expresses the relationship between the height of the fall, gravitational acceleration, and the time of fall. By rearranging this equation, we solve for \(t\), the time it takes for the car to reach the bottom of the cliff.
This concept is crucial because it implies that the rate of increase in the car's velocity is consistent, simplifying calculations and predictions about the motion. It is a foundational concept in physics that helps in understanding how objects behave when subjected to uniform gravitational forces on Earth.

Vertical and Horizontal Motion

In projectile motion, we deal with both vertical and horizontal components separately, as they behave differently. The key part of solving such problems is understanding that these components are independent.

• **Vertical Motion:** The car's vertical motion is influenced by gravity, causing it to accelerate downwards from rest, which is a classical example of uniformly accelerated motion.
• **Horizontal Motion:** The horizontal motion, however, is independent of gravity and remains constant. The car moves forward at a uniform speed, which remains unchanged until impact.

For the car's horizontal motion, we use the equation \( x = vt \) where \( x \) is the horizontal distance traveled (150 meters in this exercise), \( v \) is the constant horizontal speed, and \( t \) is the time calculated from the vertical motion.
By focusing separately on both components and using the right kinematic equations, we can effectively determine unknown quantities like the initial speed of the car when it left the cliff.