Problem 31 - Thomas Calculus

Question

Vertical motion The height of an object moving vertically is given by $$ s=-16 t^{2}+96 t+112 $$ with \(s\) in feet and \(t\) in seconds. Find a. the object's velocity when \(t=0\) b. its maximum height and when it occurs c. its velocity when \(s=0\)

Step-by-Step Solution

Answer

a. The velocity at \( t = 0 \) is 96 ft/s. b. The maximum height is 256 ft at \( t = 3 \) s. c. The velocity when \( s = 0 \) is -128 ft/s.

Step 1: Understanding the Function

The function for the height of the object is given by \[ s(t) = -16t^2 + 96t + 112 \]This is a quadratic function representing vertical motion under gravity, where \( s(t) \) is the height in feet and \( t \) is the time in seconds.

Step 2: Find the Velocity Function

The velocity of the object is the derivative of the height function \( s(t) \) with respect to time. Thus, \[ v(t) = \frac{d}{dt}[-16t^2 + 96t + 112] = -32t + 96 \] This represents the object's velocity in feet per second at any time \( t \).

Step 3: Calculate Velocity at t=0

Substitute \( t = 0 \) into the velocity function:\[ v(0) = -32(0) + 96 = 96 \] Therefore, the object's velocity at \( t = 0 \) is 96 feet per second.

Step 4: Find Maximum Height

The object reaches its maximum height at the vertex of the parabola represented by \( s(t) \). The time coordinate of the vertex is given by \[ t = -\frac{b}{2a} \] where \( a = -16 \) and \( b = 96 \):\[ t = -\frac{96}{2(-16)} = 3 \] To find the maximum height, substitute \( t = 3 \) back into the height function:\[ s(3) = -16(3)^2 + 96(3) + 112 = 256 \] The maximum height is 256 feet and it occurs at \( t = 3 \) seconds.

Step 5: Find Velocity When s=0

To find the velocity when \( s = 0 \), first solve for \( t \) when \( s(t) = 0 \): \[ -16t^2 + 96t + 112 = 0 \] Using the quadratic formula, \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -16 \), \( b = 96 \), and \( c = 112 \):\[ t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(112)}}{2(-16)} = \frac{-96 \pm \sqrt{2304 + 7168}}{-32} = \frac{-96 \pm 128}{-32} \] This simplifies to \( t = 7 \) or \( t = -1 \). Since time cannot be negative, \( t = 7 \).Substitute \( t = 7 \) into the velocity function:\[ v(7) = -32(7) + 96 = -128 \] The velocity when \( s = 0 \) is -128 feet per second.

Key Concepts

Quadratic Function

Vertical motion often involves quadratic functions that describe the trajectory of an object. In the case of our exercise, the height function is given by \[ s(t) = -16t^2 + 96t + 112 \]. Quadratic functions have a distinct parabolic shape, which can open upwards or downwards depending on the sign of the coefficient of the squared term. Here, the coefficient is \(-16\), signifying a downwards-facing parabola, indicating the object rises and then falls back down. Quadratic equations are commonly structured as \[ ax^2 + bx + c \], where \(a\), \(b\), and \(c\) are constants. In this scenario:

• \(a = -16\)
• \(b = 96\)
• \(c = 112\)

Key features include the vertex, which is the peak of the parabola in vertical motion, representing the maximum height the object reaches. By understanding these features, you can tackle a range of motion problems by identifying when and where a projectile reaches specific heights.

Velocity Calculation

To understand how fast an object is moving vertically at a specific time, we use the concept of velocity. Velocity is essentially the rate at which an object's position changes with time, and in calculus, it's the derivative of the position function. For our height function \(s(t) = -16t^2 + 96t + 112\), the derivative gives us the velocity function:\[ v(t) = \frac{d}{dt}[-16t^2 + 96t + 112] = -32t + 96 \]This equation lets us calculate the velocity instantaneously at any moment \(t\).For instance, to find the velocity at the initial moment (\(t = 0\)), plug \(t = 0\) into the velocity function:\[ v(0) = -32 \cdot 0 + 96 = 96 \]At \(t = 0\), the object is moving upwards at 96 feet per second. Understanding how velocity changes over time and using derivatives to calculate it is fundamental in physics and engineering to predict the motion of objects.

Maximum Height

The quest to find the maximum height of a vertically moving object involves determining when the object reaches the vertex of its trajectory, as represented by the peak of the parabola in the quadratic function. The vertex occurs at the point:\[ t = -\frac{b}{2a} \]Applying this to \(s(t) = -16t^2 + 96t + 112\), we use:

• \(a = -16\)
• \(b = 96\)
Plugging these into the vertex formula gives:\[ t = -\frac{96}{2(-16)} = 3 \]This tells us the object reaches its maximum height at 3 seconds. To find the maximum height itself, substitute \(t = 3\) in the original height function:\[ s(3) = -16(3)^2 + 96(3) + 112 = 256 \]Hence, the maximum height is 256 feet. This insight into the vertex not only helps find the highest point in projectile motion but also the time it takes to get there, which is crucial for planning and analysis in practical applications.