Problem 22 - Applied Calculus

Question

A grapefruit is tossed straight up with an initial velocity of \(50 \mathrm{ft} / \mathrm{sec} .\) The grapefruit is 5 feet above the ground when it is released. Its height, in feet, at time \(t\) seconds is given by $$y=-16 t^{2}+50 t+5$$ How high does it go before returning to the ground?

Step-by-Step Solution

Answer

The maximum height reached by the grapefruit is 44.0625 feet.

Step 1: Understanding the Problem

We are given the formula for the height of the grapefruit: \( y = -16t^2 + 50t + 5 \). We need to find the maximum height the grapefruit reaches before it starts falling back to the ground.

Step 2: Identify the Type of Equation

The height equation \( y = -16t^2 + 50t + 5 \) is a quadratic equation of the form \( ax^2 + bx + c \). It represents a parabola opening downwards since the coefficient of the \( t^2 \) term is negative.

Step 3: Finding the Vertex of the Parabola

The vertex form of a parabola represented by \( ax^2 + bx + c \) is found at \( t = -\frac{b}{2a} \). Here, \( a = -16 \) and \( b = 50 \). Substitute these values into the formula to find \( t \).

Step 4: Calculate Time at Max Height

Substitute \( a = -16 \) and \( b = 50 \) into the vertex formula: \[ t = -\frac{50}{2(-16)} = \frac{50}{32} = \frac{25}{16} \] seconds. This is the time at which the grapefruit reaches its maximum height.

Step 5: Calculate Maximum Height

Substitute \( t = \frac{25}{16} \) back into the height equation to find the maximum height: \[ y = -16 \left( \frac{25}{16} \right)^2 + 50 \left( \frac{25}{16} \right) + 5 \] Compute the value.

Step 6: Simplify Maximum Height

\( \frac{25}{16} = 1.5625 \), then compute: \[ y = -16(1.5625)^2 + 50(1.5625) + 5 \]. First, calculate \((1.5625)^2 = 2.44140625\), then \(-16 \times 2.44140625 = -39.0625\). Calculate \(50 \times 1.5625 = 78.125\). Sum all components: \[ y = -39.0625 + 78.125 + 5 = 44.0625 \] feet.

Key Concepts

Vertex of a Parabola

In the context of a parabola, the vertex is a key point where the graph changes direction. If a parabola opens downward, as in the case of our grapefruit problem where the height equation is given by \[ y = -16t^2 + 50t + 5 \]the vertex represents the maximum point. The formula to find the x-coordinate of the vertex in a quadratic equation like this, \( ax^2 + bx + c \), is \[t = -\frac{b}{2a}. \]

• Here, \( a = -16 \) and \( b = 50 \).
• Substitute these into the vertex formula to find the time \( t \) when the grapefruit reaches its peak height:

\[t = -\frac{50}{2(-16)} = \frac{25}{16}. \]This calculation reveals that the time at which the grapefruit is at its highest point is \( \frac{25}{16} \) seconds. The vertex not only provides the time point but is crucial for identifying the maximum height the object can reach.

Maximum Height

The maximum height is reached at the vertex of the parabola, which we identified in the previous section. To find this peak height, substitute the time \( t = \frac{25}{16} \) back into the original height equation:\[y = -16 \left( \frac{25}{16} \right)^2 + 50 \left( \frac{25}{16} \right) + 5.\]

• First, calculate \( \left( \frac{25}{16} \right)^2 = 2.44140625 \).
• Then, multiply by -16: \(-16 \times 2.44140625 = -39.0625 \).
• Next, calculate \( 50 \times \frac{25}{16} = 78.125 \).
• Add all parts together: \(-39.0625 + 78.125 + 5 \).
• This results in a maximum height of \( 44.0625 \) feet.

Using this method confirms that the grapefruit reaches its peak height of approximately 44.06 feet before it begins its descent back to the ground.

Parabolic Motion

Parabolic motion describes the trajectory of any object thrown into the air with an initial velocity, not just vertically but potentially horizontally as well. In the case of the grapefruit, which is tossed straight up, the movement is a simple example of vertical parabolic motion. The path taken by the grapefruit is a symmetrical arc, indicative of a downward-opening parabola that is defined by its quadratic equation:\[y = -16t^2 + 50t + 5.\]

• "-16" signifies the force of gravity pulling the object back down to Earth.
• The initial velocity is \( 50 \) feet per second, giving it an upward thrust.
• The constant \( 5 \) represents the starting height above the ground.

This combination creates a parabolic path, achieving the highest point at the vertex, before gravity overtakes the initial velocity and the object falls. Understanding parabolic motion helps us not only to calculate maximum height and time of flight but also to appreciate the forces interacting with moving objects in our daily environment.