Question

A car is driven 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 initial velocity of the car is approximately 42.98 m/s.

Step-by-step solution

Identify relevant information

- The height of the cliff: \(h = 60.0 \mathrm{~m}\) - Distance from the base of the cliff: \(d = 150 . \mathrm{m}\) - Vertical acceleration due to gravity: \(a_y = -9.81 \mathrm{~m/s^2}\) (negative as it goes downward) - Initial vertical velocity: \(v_{0_y} = 0\) (since the car goes straight) - Initial horizontal velocity: \(v_{0_x}\) (unknown; we need to find this)

Time calculation

Since both horizontal and vertical motion share the same time duration, we can use the vertical information to find the time it takes for the car to hit the ground. Using the second equation of motion: \(h = v_{0_y}t + \frac{1}{2}a_yt^2\) Plugging in the vertical values: \(60.0 \mathrm{~m} = 0 + \frac{1}{2}(-9.81 \mathrm{~m/s^2})t^2\) Now, we solve for \(t^2\), and then \(t\). \(t^2 = \frac{2\cdot 60.0 \mathrm{~m}}{-9.81 \mathrm{~m/s^2}} \approx 12.19 \mathrm{~s^2}\) Taking the square root, we get: \(t \approx \sqrt{12.19 \mathrm{~s^2}} \approx 3.49 \mathrm{~s}\)

Calculate initial horizontal velocity

Now that we have the time when the car hits the ground, we can use the constant velocity equation for horizontal motion: \(d = v_{0_x}\cdot t\) Inserting the values: \(150 . \mathrm{m} = v_{0_x} \cdot 3.49 \mathrm{~s}\) Solving for \(v_{0_x}\): \(v_{0_x} = \frac{150 . \mathrm{m}}{3.49 \mathrm{~s}} \approx 42.98 \mathrm{~m/s}\)

Conclusion

The car was traveling at approximately \(42.98 \mathrm{~m/s}\) when it went over the cliff.

Key concepts

Understanding Kinematics Equations

Kinematics equations are the foundation of analyzing motion in physics. These equations relate an object's position, velocity, acceleration, and time without considering the forces that cause such motion. For projectile motion, there are two components to consider: horizontal and vertical. The horizontal motion is consistent with no acceleration (assuming air resistance is negligible), while the vertical motion is affected by gravity and can be described by the kinematics equations.

A key point to remember is that the vertical motion is under constant acceleration due to gravity, which leads to a parabolic trajectory when we consider both vertical and horizontal displacements. One of the most commonly used kinematics equations for vertical motion is \[ h = v_{0_y}t + \frac{1}{2}a_yt^2 \.\] This very equation allows us to calculate important variables such as displacement (\(h\)), initial velocity (\(v_{0_y}\)), time (\(t\)), and acceleration (\(a_y\)). By rearranging and solving this equation as demonstrated in the original exercise's Step 2 solution, we determine the time it took for the car to fall to the ground after plummeting from the cliff.

Explaining Free Fall Acceleration

Free fall acceleration is a key concept when understanding projectile motion. On Earth, any object in free fall (without the effects of air resistance) will accelerate downwards due to gravity at approximately \[ a_y = -9.81 \mathrm{~m/s^2} \.\] This is a constant value, which means that irrespective of the object's mass, it will gain speed at this rate when falling.

In the context of our example with the car, we assume the only force acting on the car in the vertical direction is gravity, causing it to accelerate at this rate towards the Earth. Free fall acceleration is negative in the equation because we take downwards as the negative direction. This helps in calculating the time of flight and is critical to finding the horizontal velocity, which is our ultimate goal in the provided problem. It's the reason why the car's initial vertical velocity is zero and progressively increases in magnitude as the car descends.

Determining Horizontal Velocity Calculation

In projectile motion problems, calculating horizontal velocity is crucial for understanding how far an object travels before hitting the ground. Since we often assume no horizontal acceleration (neglecting air resistance), the horizontal velocity remains constant throughout the projectile's flight.

The equation used to determine it is usually the definition of velocity: \[ v_x = \frac{d}{t} \.\] where \(v_x\) is the horizontal velocity, \(d\) is the horizontal distance traveled, and \(t\) is the time of flight. This simplifies our calculations as we do not need to consider horizontal acceleration. For the car in our scenario, once we have the time it spent in the air with the help of kinematics equations and free fall acceleration, we can easily compute its horizontal velocity by rearranging the horizontal velocity formula (Step 3 in the original solution). This gives us the speed at which the car was traveling horizontally when it went over the edge of the cliff.