Problem 50 - Beginning and Intermediate Algebra

Question

A projectile on Earth is fired straight upward so that its distance (in feet) above the ground \(t\) seconds after firing is given by$$s(t)=-16 t^{2}+400 t$$Find the maximum height it reaches and the number of seconds it takes to reach that height.

Step-by-Step Solution

Answer

The maximum height is 2500 feet, reached in 12.5 seconds.

Step 1: Identify the equation

The height of the projectile as a function of time is given by the equation \[ s(t) = -16t^2 + 400t \]

Step 2: Determine the type of function

The equation \[ s(t) = -16t^2 + 400t \] represents a quadratic function in the form \[ at^2 + bt + c \] with coefficients \[ a = -16 \] and \[ b = 400 \], where \[ c = 0 \]. This is a downward parabola.

Step 3: Find the vertex (maximum height)

The maximum height of a projectile in a quadratic equation \[ at^2 + bt + c \] occurs at \[ t = -\frac{b}{2a} \]. For \[ a = -16 \] and \[ b = 400 \], we calculate \[ t = -\frac{400}{2(-16)} = \frac{400}{32} = 12.5 \] seconds.

Step 4: Calculate the maximum height

Substitute \[ t = 12.5 \] back into the height equation: \[ s(12.5) = -16(12.5)^2 + 400(12.5) = -16(156.25) + 5000 = -2500 + 5000 = 2500 \] feet.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function where the highest degree of the variable is 2. This means the general form of a quadratic function is given by \[ ax^2 + bx + c \]. Here, the coefficients are: \( a \): determines the direction and width of the parabola, \( b \): influences the position of the parabola along the x-axis, \( c \): represents the y-intercept. In our problem, the quadratic function for the projectile's height is \[ s(t) = -16t^2 + 400t \]. Quadratic functions create parabolic graphs, which can either open upwards or downwards depending on the sign of \( a \). If \( a \) is negative, like in the given problem where \( a = -16 \), the parabola opens downward.

Maximum Height

To find the maximum height of the projectile, we need to understand the properties of a downward parabola. The maximum height corresponds to the highest point on the graph of the quadratic function, also known as the vertex. The vertex of a parabola opening downward is the highest point. We use the formula to find the time \( t \) when this maximum height is reached: \[ t = -\frac{b}{2a} \]. For our function \[ s(t) = -16t^2 + 400t \], the coefficients are \( a = -16 \) and \( b = 400 \). Substituting these values, we get: \[ t = -\frac{400}{2(-16)} = \frac{400}{32} = 12.5 \] seconds. This means the projectile reaches its maximum height 12.5 seconds after being fired.

Vertex of a Parabola

The vertex of a parabola is a significant point that represents either the maximum or minimum value of the quadratic function, depending on whether it opens downward or upward. For the downward parabola seen in our problem, the vertex is the maximum point. From our quadratic function \[ s(t) = -16t^2 + 400t \], we identified the vertex's x-coordinate (time in this context) as \( t = 12.5 \) seconds. To find the y-coordinate (height), we substitute \( t = 12.5 \) back into the equation: \[ s(12.5) = -16(12.5)^2 + 400(12.5) = -16(156.25) + 5000 = -2500 + 5000 = 2500 \]. So, the vertex of the parabola is at \( (12.5, 2500) \), representing that the projectile's maximum height is 2500 feet.

Downward Parabola

A downward parabola is a curve that opens downwards, meaning it has a maximum point but no minimum point. This is common in projectile motion problems where the object rises to a peak height and then falls back down. The equation \[ s(t) = -16t^2 + 400t \] is an example of a downward parabola since it has a negative coefficient (\( a = -16 \)) for the \( t^2 \) term. This affects the graph by curving it downward. The term \( -16t^2 \) pulls the graph down as \( t \) increases, while \( 400t \) initially pulls it upwards until the peak point is reached and then gravity brings it back down. Understanding the nature of a downward parabola helps solve projectile motion problems efficiently.