Question

A projectile is fired straight upward from ground level with an initial velocity of 200 feet per second. During what time period will its height exceed 400 feet?

Short answer

The time during which the projectile's height exceeds 400 feet is from \( t = 5 s \) to \( t = 25 s \).

Step-by-step solution

Find time when height is 400 feet

The equation of height \( h \) is given as \( h = ut - \frac{1}{2}gt^2 \). Substituting the values \( u = 200 ft/s \), \( g = -32 ft/s^2 \) (because the motion is upwards), and \( h = 400 ft \), and then solving for \( t \), we obtain two values: \( t = 5 s \) and \( t = 25 s \). These represent the times when the ball reaches 400 feet while going up and coming down respectively.

Determine time period

So, the ball will first reach 400 feet at \( t = 5 s \) and then will come down to 400 feet again at \( t = 25 s \). Hence, the ball exceeds 400 feet in the time period between \( t = 5 s \) and \( t = 25 s \).

Key concepts

Quadratic Equations

Quadratic equations are fundamental to understanding many concepts in algebra and are frequently utilized in physics problems, including projectile motion. A quadratic equation is any equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \) and \( c \) are constants, and \( a \eq 0 \). In our exercise, the equation of the projectile motion is a real-world example of a quadratic equation where the height \( h \) changes with time \( t \) and can be written as \( h = ut - \frac{1}{2}gt^2 \).

The process of solving a quadratic equation is crucial because it can yield two solutions, representing two distinct events in time—in our exercise, these times correspond to the moments when the projectile reaches a certain height twice; once while going up and again while coming down. To find these moments, one would usually rearrange the equation to a standard form and apply methods such as factoring, completing the square, or using the quadratic formula. The resulting two solutions help determine the important times related to the projectile's flight.

Physics in Algebra

When integrating physics and algebra, we often translate physical phenomena into mathematical equations. In projectile motion problems, for instance, we use algebra to describe the motion of the projectile through space over time. In our exercise, we used physical constants: the initial velocity \( u \) and the acceleration due to gravity \( g \).

The significance of algebra in physics is its ability to simplify and solve complex problems by representing them with equations. This requires an understanding of how to manipulate equations and variables to isolate and solve for the unknowns, as we do when we solve for \( t \) (time) given the initial conditions. Algebraic techniques enable us to predict the behavior of physical systems, such as calculating when a projectile reaches a certain height and predict its future positions or velocities at various times.

Initial Velocity

Initial velocity, denoted as \( u \) in equations, is the velocity with which an object starts its motion. In our projectile problem, the initial velocity is given as 200 feet per second. This value is pivotal in solving the equation as it represents the projectile's speed at the moment of launch.

Understanding initial velocity is necessary because in projectile motion, it influences the maximum height reached and the duration of the flight. The upward motion against gravity slows down due to the acceleration due to gravity \( g \), until velocity momentarily drops to zero at the peak of the trajectory. After that, gravity accelerates the object back toward the ground. The interplay between initial velocity and gravitational acceleration determines the time instants when the projectile reaches specific heights. The two times we found in the solution represent the upward and downward crossings of the 400-feet mark, showcasing how initial velocity sets the stage for the entire motion of the projectile.