Question

Neglecting air resistance, the upward velocity of the water in the stream of a particular fountain is given by the formula \(v=-32 t+28\), where \(t\) is the number of seconds after the water leaves the fountain. While going upward, the water slows down until, at the top of the stream, the water has a velocity of 0 feet per second. How long does it take a droplet of water to reach the maximum height?

Short answer

The droplet reaches the maximum height after \(7 / 8\) or 0.875 seconds.

Step-by-step solution

Understand the given equation

In the given equation \(v = -32 t + 28\), the velocity of the water droplet changes over time due to the effect of gravity (represented by -32). The initial velocity is given by the constant term, i.e., 28 feet per second. Since we know at the highest point, the velocity of droplet will be 0. This fact can be used to find the required time.

Set the velocity to zero

To find when the droplet reaches the maximum height, set the velocity \(v\) to 0 in the given equation. This gives: 0 = -32t + 28.

Solve for time

Rearrange the equation in Step 2 and solve for \(t\): Add 32t to both sides of the equation to isolate the term containing \(t\) on one side: 32t = 28. Then divide both side of the equation by 32 to compute for \(t\): \(t = 28 / 32\).

Simplify the result

The fraction \(t = 28 / 32\) simplifies by dividing both the numerator and the denominator by 4, which gives: \(t = 7 / 8\).

Key concepts

Understanding Quadratic Equations

Quadratic equations are a fundamental aspect of algebra and frequently occur in various scientific calculations, including motion problems. These equations can be recognized by their standard form, which is \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). In the context of motion, the variable \(x\) is often replaced with time \(t\), and the coefficients relate to different factors affecting the motion, such as initial velocity and acceleration due to gravity.
In the fountain problem, the equation \(v = -32t + 28\) may not look like a standard quadratic equation because it is missing the \(t^2\) term. However, it is derived from the quadratic position function, where the equation represents the velocity \(v\) of water at time \(t\) after considering gravity's acceleration. Here, \( -32 \equiv a\) represents gravity's acceleration in feet per second squared, and \( 28 \equiv c\) represents the initial velocity in feet per second. Solving quadratic equations typically involves finding the values of \(x\) (or \(t\) in this case) that make the equation true, which corresponds to significant points in the motion, such as when the water reaches the fountain's maximum height.

Calculating Velocity and Time

Velocity and time are inextricably linked in motion problems. In physics, velocity is the rate of change of an object's position with respect to time. It is a vector quantity, which means it has both magnitude (speed) and direction. The equation from the exercise \(v = -32t + 28\) is a velocity-time relationship that describes how the velocity of the water changes as time progresses.
When solving for time, as in the fountain problem, you're essentially finding the moment when an event happens, such as reaching the maximum height, which occurs when velocity is zero. Setting the velocity to zero in the given equation transforms the problem into finding the root of the linear function. This instance doesn't produce a quadratic equation since the \(t^2\) term is absent; however, the constant acceleration due to gravity lays the foundation for a scenario typically governed by quadratic motion equations.

Gravity's Effect on Motion

Gravity significantly influences motion problems on Earth, typically contributing a constant acceleration downward of about \(9.8 \text{m/s}^2\) or \(32 \text{ft/s}^2\), depending on the unit system used. This acceleration is what changes the velocity of an object over time, and it is represented by the coefficient of the \(t\) term in our velocity equation.
In this scenario, without the impact of air resistance, the quadratic nature of gravitational motion ensures that the water's upward journey through the air can be modeled predictably, with the velocity decreasing linearly until it becomes zero at the peak. The time it takes to reach this point is calculated by setting the velocity equal to zero and solving for \(t\), which reflects the time it takes for gravity to decelerate the water droplet to a standstill. Understanding this effect is crucial in many fields, including engineering and aerospace, where predicting the trajectory of projectiles is necessary.