Question

You drop a rock over the edge of a cliff from a height \(h\). Your friend throws a rock over the edge from the same height with a speed \(v_{0}\) vertically downward, at some time \(t\) after you drop your rock. Both rocks hit the ground at the same time. How long after you dropped your rock did your friend throw hers? Express your answer in terms of \(v_{0}, g,\) and \(h\).

Short answer

Answer: The time after the first rock is dropped when the second rock is thrown can be expressed as \(t = \frac{v_{0} T - \frac{1}{2} g T^2}{v_{0} - g T}\), where \(v_0\) is the initial velocity of the second rock, \(g\) is the acceleration due to gravity, and \(T\) is the total time both rocks take to reach the ground.

Step-by-step solution

Analyze the motion of the first rock

First, we need to analyze the motion of the first rock that was dropped from the cliff. The rock is freely falling, which means its initial velocity is 0 (\(v_{0} =0\)). Since it is falling, we will consider the downward direction as positive, and the acceleration due to gravity as positive \(g\). The displacement is equal to the height \(h\) of the cliff.

Determine the equation of motion for the first rock

Next, we will find an equation of motion that involves the displacement, initial velocity, acceleration, and time for the first rock. Using this equation, we can calculate the time it takes for the first rock to fall and hit the ground: \(h = v_{0} t_{1} + \frac{1}{2} g t_{1}^2\) Since the initial velocity is 0, this simplifies to: \(h = \frac{1}{2} g t_{1}^2\)

Analyze the motion of the second rock

Now, we need to analyze the motion of the second rock that was thrown vertically downward with an initial velocity \(v_0\). The displacement is also equal to the height of the cliff \(h\). The acceleration is due to gravity \(g\), and the second rock took a time of \((t_2 - t)\) to hit the ground.

Determine the equation of motion for the second rock

Next, we will find an equation of motion that involves the displacement, initial velocity, acceleration, and time for the second rock. Using this equation, we can calculate the time it takes for the second rock to fall and hit the ground: \(h = v_{0}(t_2 - t) + \frac{1}{2} g (t_2 - t)^2\) This is the equation of motion of the second rock.

Set equal times for both rocks and solve for \(t\)

Since both rocks hit the ground at the same time, we can set their times equal to each other: \(t_1 = t_2 - t\) Now, we can substitute the equation of motion for the first rock into this equation, using the fact that \(h = \frac{1}{2} g t_{1}^2\): \(\frac{1}{2} g t_{1}^2 = v_{0}(t_2 - t) + \frac{1}{2} g (t_2 - t)^2\) Now, we want to solve for the time \(t\) which is the time after the first rock was dropped when the second rock is thrown. Rearranging the equation to isolate \(t\): \(\frac{1}{2}g(t_2^2 - 2 t t_2 + t^2) - v_{0} t + v_{0} t_2 = 0\) For not getting too messy, let's call \(t_2\) as the total time both rocks take to reach the ground and call it \(T\). Since they fall simultaneously, rewrite the previous equation as: \(\frac{1}{2}g(T^2 - 2 t T + t^2) - v_{0} t + v_{0} T = 0\) Divide the equation by \(g\): \(\frac{1}{2}(T^2 - 2 t T + t^2) - \frac{v_{0}}{g} t + \frac{v_{0}}{g} T = 0\)

Express \(t\) in terms of \(v_0, g,\) and \(h\)

To find \(t\), we can now solve the above equation for the time \(t\) in terms of \(v_0, g,\) and \(h\). In this way, the time it takes for the second rock to reach the ground can be expressed in terms of the initial velocity, acceleration, and height of the cliff: \(t = \frac{v_{0} T - \frac{1}{2} g T^2}{v_{0} - g T}\) The calculated value of \(t\) represents the time after the first rock was dropped when the second rock was thrown.

Key concepts

Free Fall Motion

Free fall motion refers to the movement of an object under the influence of gravity alone, with no other forces acting upon it, such as air resistance. It's crucial to understand that all objects in free fall, regardless of their mass, accelerate towards the ground at the same rate when in a vacuum, like the rocks in our problem. This is a result of the acceleration due to gravity.

For objects in free fall, their motion can be described using kinematic equations. These equations relate initial velocity, final velocity, acceleration, time, and displacement. When an object is simply dropped, like the first rock in our exercise, its initial velocity (\(v_{0}\)) is zero, and it accelerates downward due to gravity.

Equations of Motion

The equations of motion are a set of formulas that predict the future position and velocity of an object moving under constant acceleration. These are critically important in kinematics problems, as they allow us to describe the trajectory of an object, such as our rocks being dropped from a cliff.

One key equation used when analyzing objects in free fall is \( s = ut + \frac{1}{2}at^2 \), where \(s\) is the displacement, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. In free fall motion, \(s\) corresponds to the change in height (\(h\)), \(u\) is often zero if the object is dropped, and \(a\) is the acceleration due to gravity (\(g\)). By adjusting the equation to fit specific scenarios, like the late throw of the second rock in our problem, we can solve complex kinematics problems.

Acceleration Due to Gravity

Acceleration due to gravity, denoted as \(g\), is the rate at which objects accelerate towards the Earth when they are in free fall. This value is approximately \(9.8 \text{ m/s}^2\) near the Earth's surface, meaning for every second an object is in free fall, its velocity increases by \(9.8 \text{ m/s}\).

In kinematic problems involving free fall, such as our textbook scenario, \(g\) is a critical factor. It is because of this constant acceleration that an object dropped from rest will increase its velocity as it falls, and why an object projected downward will have its initial velocity \(v_0\) incremented by \(9.8 \text{ m/s}\) for every second it is in the air. The universality of \(g\) allows us to solve for various unknowns in kinematic equations by incorporating the known value of gravitational acceleration.