Question

In Exercises 71-76, assume that air resistance is negligible, which implies that the position equation \(s=-16 t^{2}+v_{0} t+s_{0}\) is a reasonable model Falling Object A rock is dropped from the top of a 200 -foot cliff that overlooks the ocean. How long will it take for the rock to hit the water?

Short answer

The rock will hit the water after approximately \( \sqrt{200/16} \) seconds, which equals to approximately 3.54 seconds.

Step-by-step solution

Identifying the known parameters

From the problem, we know that the initial velocity \(v_{0} = 0\) m/s (because the rock is dropped and not thrown), the initial height \(s_{0} = 200\) feet, and the final height \(s\) is 0 (as it hits the water surface).

Substituting the known parameters into the position equation

The problem involves applying the kinematic equation \(s=-16 t^{2}+v_{0} t+s_{0}\). Substituting the knowns into the equation, we get \(0 = -16t^2 + 0 * t + 200\). Since the initial velocity \(v_{0}\) of the rock is 0, the term \(v_{0} * t\) cancels out, simplifying our equation to \(0 = -16t^2 + 200\).

Solve the equation for time \(t\)

To solve for \(t\), girst move \(16t^2\) to the other side, so the equation becomes \(16t^2 = 200\). Then, by taking the square root on both side, \(t\) equals to \(sqrt(200/16)\).

Key concepts

Free Fall Motion

Understanding free fall motion is essential to solve problems involving objects dropped from a height, like in our problem concerning a rock being dropped from a cliff. Free fall motion occurs when the only force acting on an object is gravity, implying there is no air resistance. This condition can be simplified with the acceleration due to gravity, often denoted as 'g' and approximately equals to \(9.8 m/s^2\) or \(32 ft/s^2\) in imperial units.

When dealing with free fall problems, we assume that the acceleration of the object is constant and directed downwards, towards the center of the Earth. This acceleration remains the same throughout the motion until the object hits the ground. It's crucial to understand that during free fall, the only velocity the object gains comes from gravity's pull, which impacts the overall time the object takes to fall and its final velocity upon impact.

Solving Quadratic Equations

Quadratic equations come up frequently in physics, especially in kinematics. They have the general form \(ax^2 + bx + c = 0\), where 'a', 'b', and 'c' are constants. In our exercise, the quadratic equation appears after inputting the known values into the position equation, simplifying to \(16t^2 = 200\).

To solve for 't', we need to isolate it on one side of the equation. This often involves dividing by the coefficient of \(t^2\), and then finding the square root of both sides. Remember to consider both positive and negative roots, but in the case of time, only the positive root makes physical sense. The final answer should be realistic and correspond with the context of the problem—here, indicating the time it takes for the rock to reach the water surface after being dropped.

Initial Velocity

In kinematic problems, the initial velocity, denoted as \(v_0\), is the velocity at which an object starts its motion. It is a vector quantity with both magnitude and direction. For free fall problems where the object starts from rest, as with the rock in our example, the initial velocity is zero. This is critical because it simplifies the quadratic equation, eliminating the \(v_0 t\) term.

When the initial velocity is not zero, it must be factored into the calculations, especially if the object is thrown upwards or downwards with an initial force. Remember that the direction of the initial velocity affects the motion's trajectory and overall time to reach a certain position.

Position Equation

The position equation for free fall motion \(s=-16 t^{2}+v_{0} t+s_{0}\) describes the object's position over time. In this equation, 's' is the final position, \(t\) is the time elapsed, \(v_0\) is the initial velocity, and \(s_0\) is the initial position or height from which the object is dropped.

When an object is dropped, and air resistance is negligible, the position equation simplifies as the initial velocity is zero, and the only acceleration is due to gravity. In our scenario with the rock, known values such as the initial height of 200 feet are plugged into this equation to solve for the time 't'. This equation is significant in physics because it directly relates an object's movement under constant acceleration to time, which is a commonly analyzed scenario in kinematic problems.