Question

Impact speed A rock is dropped off the edge of a cliff, and its distance \(s\) (in feet) from the top of the cliff after \(t\) seconds is \(s(t)=16 t^{2} .\) Assume the distance from the top of the cliff to the ground is \(96 \mathrm{ft}\) a. When will the rock strike the ground? b. Make a table of average velocities and approximate the velocity at which the rock strikes the ground.

Short answer

Question: Estimate the velocity at which a rock strikes the ground when dropped from a height of 96 feet, using the equation s(t) = 16t^2 and a table of average velocities. Answer: The rock strikes the ground at an approximate velocity of 49.44 ft/s.

Step-by-step solution

Find the time it takes for the rock to strike the ground.

Given the equation for distance \(s(t) = 16t^2\), and the distance from the top of the cliff to the ground is \(96ft\), we need to find the time it takes for the rock to reach the ground. To do this, we can set the distance \(s(t)\) equal to \(96\): \(96 = 16t^2\) Now we just need to solve for t: \(t^2 = \frac{96}{16}\) \(t^2 = 6\) \(t = \sqrt{6}\) Since time cannot be negative, the time it takes for the rock to strike the ground is \(t = \sqrt{6}\) seconds.

Make a table of average velocities

To find the average velocity, we need to find the difference in distances divided by the difference in time intervals. Let's create a table for average velocities for a few time intervals close to when the rock strikes the ground: Time \((t_i)\) | Time \((t_f)\) | Distance \((s_i)\) | Distance \((s_f)\) | Average velocity \((v_i)\) --- | --- | --- | --- | --- \(\sqrt{6}-0.1\) | \(\sqrt{6}\) | \(16(\sqrt{6}-0.1)^2\) | \(16\sqrt{6}^2\) | \(\frac{16(\sqrt{6}^2) - 16(\sqrt{6}-0.1)^2}{\sqrt{6}-(\sqrt{6}-0.1)}\) \(\sqrt{6}-0.01\) | \(\sqrt{6}\) | \(16(\sqrt{6}-0.01)^2\) | \(16\sqrt{6}^2\) | \(\frac{16(\sqrt{6}^2) - 16(\sqrt{6}-0.01)^2}{\sqrt{6}-(\sqrt{6}-0.01)}\) Now let's calculate the average velocities for these time intervals: \(v_1 = \frac{16(\sqrt{6}^2) - 16(\sqrt{6}-0.1)^2}{\sqrt{6}-(\sqrt{6}-0.1)}\) \(v_1 \approx 48.96\,\text{ft/s}\) \(v_2 = \frac{16(\sqrt{6}^2) - 16(\sqrt{6}-0.01)^2}{\sqrt{6}-(\sqrt{6}-0.01)}\) \(v_2 \approx 49.92\,\text{ft/s}\)

Approximate the velocity at which the rock strikes the ground.

Now we have two average velocities for time intervals very close to when the rock strikes the ground. We can use these values to approximate the velocity at which the rock strikes the ground. An estimate for the velocity can be obtained by taking the average of the two calculated average velocities: \(v \approx \frac{48.96 + 49.92}{2}\) \(v \approx 49.44\,\text{ft/s}\) Based on the table of average velocities, we can approximate the velocity at which the rock strikes the ground as \(49.44\,\text{ft/s}\).

Key concepts

Projectile Motion

Projectile motion is a form of motion experienced by an object thrown near the Earth's surface and moving along a curved path under the action of gravity only. This concept is essential in solving many real-world problems in physics, engineering, and sports. When a rock is dropped from a cliff, as in our exercise, it is essentially a projectile undergoing vertical motion under the influence of gravity.

Understanding projectile motion involves analyzing two independent components: horizontal and vertical motions. Since our rock is dropped straight down, there is no horizontal component, and we focus solely on the vertical free fall aspect. The trajectory of the projectile, which in the absence of air resistance would form a parabola in cases involving horizontal displacement, is not applicable here. Yet, the principles governing the vertical component are identical to those of any other projectile motion.

Velocity Approximation

In kinematics, determining the exact velocity of a moving object at a particular instant can be challenging without calculus. To resolve this, velocity approximation methods are utilized to estimate the object's speed. In our exercise, average velocities at different intervals give us an estimation of the impact velocity of the rock.

To better understand the steps provided in the solution, we calculate average velocities over progressively shorter time intervals leading up to the moment of impact. By doing this, we approach the instantaneous velocity, or the true velocity, of the rock at the moment it hits the ground. As the intervals become smaller, our approximation becomes more precise, highlighting how differential calculus can be used to find exact velocities if required.

Free Fall

Free fall occurs when an object is moving solely under the force of gravity, with no other forces acting on it (like air resistance). For objects in free fall near Earth's surface, they accelerate downwards at a constant rate: the standard gravitational acceleration, which is approximately 9.8 meters per second squared (\(9.8 \text{m/s}^2\) or, in our exercise, using imperial units, 32 feet per second squared).The position of an object in free fall can be described by the equation introduced in the exercise, \(s(t) = 16t^2\). This equation is a simplified version of the fundamental kinematics equation for vertical motion under constant acceleration, tailored for the specific acceleration due to gravity here on Earth (using 16 feet per second squared as an approximation). In our problem, we derived the time it takes for the rock to hit the ground from this equation and subsequently worked out the velocity using intervals close to the point of impact.