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 difference between the two drops is \(t = t_1 - \frac{2v_0}{g}\), where \(t_1 = \sqrt{\frac{2h}{g}}\) is the time it takes for the first rock to hit the ground.

Step-by-step solution

Set up equations for time it takes for both rocks to hit the ground

We'll use the equation \(y = y_{0} + v_{0}t + \frac{1}{2}at^2\), where \(y\) is the final position, \(y_{0}\) is the initial position, \(v_{0}\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration. Since both rocks start at the same height \(h\) and hit the ground at the same time, we can set their final positions to 0. The equations for the time for both rocks are: For the first rock: \(0 = h + 0 \cdot t_1 - \frac{1}{2}g t_1^2\) For the second rock: \(0 = h + v_{0}(t_1 - t) - \frac{1}{2}g (t_1 - t)^2\)

Solve the equations for time

First, solve the equation for the first rock, \(0 = h - \frac{1}{2}g t_1^2\). Then, we have \(t_1^2 = \frac{2h}{g}\). Let's call this equation (1). Now, let's solve the equation for the second rock. We can rewrite the equation as \(0 = h + v_{0}(t_1 - t) - \frac{1}{2}g(t_1^2 - 2t_1t + t^2)\). We know \(t_1^2 = \frac{2h}{g}\) from equation (1), so substitute this into the second equation. We get: \(0 = h + v_{0}(t_1 - t) - \frac{1}{2}g \left(\frac{2h}{g} - 2t_1t + t^2 \right)\)

Simplify and find the time difference

The equation becomes: \(0 = h + v_{0}(t_1 - t) - (h - t_1gt + \frac{1}{2}gt^2)\). Simplify and solve for \(t\), the time difference between the two drops: \(t(2v_0 - gt) = 2v_0t_1\). Divide both sides by \(2t\), we obtain: \(v_0 - \frac{1}{2}gt = v_0t_1\) Thus, the time difference between the two drops is: \(t = t_1 - \frac{2v_0}{g}\) Where \(t_1 = \sqrt{\frac{2h}{g}}\) is the time it takes for the first rock to hit the ground.

Key concepts

Equations of Motion

Understanding the equations of motion is crucial when solving kinematics problems, particularly those involving objects moving under the influence of gravity. In physics, these equations describe the relationship between an object's displacement, initial velocity, acceleration, and the time elapsed. The general form of the equations is given by:

• Displacement: \( s = ut + \frac{1}{2}at^2 \),
• Final velocity: \( v = u + at \),
• Final velocity squared: \( v^2 = u^2 + 2as \),
• Displacement after a time: \( s = \frac{u + v}{2} \cdot t \).

Where:

• \( s \) is the displacement,
• \( u \) is the initial velocity,
• \( v \) is the final velocity,
• \( t \) is the time elapsed, and
• \( a \) is the acceleration.

For vertically thrown objects, such as the rocks in our problem, the motion is usually described using the first formula, where displacement is interpreted as height \( h \), and acceleration \( a \) is due to gravity, usually denoted as \( g \). By manipulating these equations, we can determine various quantities of interest, such as the time it takes for an object to reach the ground.

Free Fall

Free fall is a motion where gravity is the only force acting on an object. When discussing free fall, we often neglect air resistance, leading to a constant acceleration towards the center of the Earth, known as the acceleration due to gravity. In free fall problems, objects are either dropped, thrown upwards, or thrown downwards. The time it takes for such an object to reach the ground depends entirely on its initial height and the acceleration due to gravity.

An object in free fall has an initial velocity of zero if it is simply dropped, as with the first rock in the problem. However, if an object is thrown downwards, like the second rock, it has an initial velocity that must be accounted for when analyzing its motion.

Acceleration due to Gravity

Earth's gravity causes all objects to accelerate towards its center at approximately \( 9.81 \) meters per second squared \( (m/s^2) \). This acceleration is denoted as \( g \) in equations. When objects fall toward Earth in a vacuum, or if we decide to ignore air resistance, this acceleration remains constant. The influence of gravity is essential in our kinematics problem because it's the force that causes both rocks to accelerate downwards regardless of their initial velocities.

Knowing the constant acceleration due to gravity allows us to use the equations of motion to predict the behavior of objects in free fall, not only for when they will reach the ground but also for other parameters such as their velocity at any point during the fall.

Initial Velocity

The term 'initial velocity' denotes an object's velocity at the start of a time period under consideration. Initial velocity is critical in kinematics problems and has a direct influence on subsequent motion. If an object is dropped, its initial velocity is zero. However, if an object is thrown, its initial velocity would be the throwing speed, which can significantly alter the trajectory and timing of its motion.

In our problem, knowing the initial velocity \( v_{0} \) of the second rock thrown downward is vital for determining how long it takes for both rocks to hit the ground at the same time. Objects thrown downward with an initial velocity reach the ground quicker than those dropped. As the solution shows, a comparison between times taken by objects with different initial velocities allows us to deduce the time delay between their releases.