Problem 89 - Algebra and Trigonometry

Question

Use the formula \(h=-16 t^{2}+v_{0} t\) discussed in Example 7. A ball is thrown straight upward at an initial speed of \(v_{0}=40 \mathrm{ft} / \mathrm{s}\). (a) When does the ball reach a height of 24 \(\mathrm{ft}\) ? (b) When does it reach a height of 48 \(\mathrm{ft}\) ? (c) What is the greatest height reached by the ball? (d) When does the ball reach the highest point of its path? (e) When does the ball hit the ground?

Step-by-Step Solution

Answer

(a) 0.5s, 3s; (b) 0.75s, 4s; (c) 25 ft; (d) 1.25s; (e) 2.5s.

Step 1: Set up the equation for part (a)

For finding when the ball reaches a height of 24 ft, we substitute the height into the formula: \[ h = -16t^2 + v_0 t \]For this problem, set \( h = 24 \) and \( v_0 = 40 \):\[ 24 = -16t^2 + 40t \].

Step 2: Rearrange and solve the equation for part (a)

Rearrange to set the equation to zero:\[ -16t^2 + 40t - 24 = 0 \].Use the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16 \), \( b = 40 \), and \( c = -24 \). Calculate \( t \):\[ t = \frac{-40 \pm \sqrt{40^2 - 4(-16)(-24)}}{2(-16)} \].This simplifies to: \( t = 3, 0.5 \).

Step 3: Set up the equation for part (b)

For finding when the ball reaches a height of 48 ft, substitute the height into the formula:\[ h = -16t^2 + 40t \].Set \( h = 48 \):\[ 48 = -16t^2 + 40t \].

Step 4: Rearrange and solve the equation for part (b)

Rearrange to set the equation to zero:\[ -16t^2 + 40t - 48 = 0 \].Use the quadratic formula where \( a = -16 \), \( b = 40 \), and \( c = -48 \):\[ t = \frac{-40 \pm \sqrt{40^2 - 4(-16)(-48)}}{2(-16)} \].This simplifies to: \( t = 4, 0.75 \).

Step 5: Find greatest height for part (c)

The greatest height is reached at the vertex of the parabola. For the equation \(-16t^2 + v_0t\), the formula for vertex time \(t\) is \( t = \frac{-b}{2a} \), where \( a = -16 \) and \( b = 40 \):\[ t = \frac{-40}{2(-16)} = 1.25 \].Thus, substitute back for height: \( h = -16(1.25)^2 + 40(1.25) \).Calculate \( h \):\[ h = 25 \].

Step 6: Highest point time for part (d)

From Step 5, the ball reaches its highest point at \( t = 1.25 \) seconds.

Step 7: When the ball hits the ground for part (e)

Set \( h = 0 \) in the original equation for when the ball hits the ground:\[ 0 = -16t^2 + 40t \].Factor out \( t \):\[ t(-16t + 40) = 0 \].Solve for \( t \):\( t = 0 \) (initial moment) or \( t = \frac{40}{16} = 2.5 \).

Key Concepts

Parabolic Motion

The motion of a ball thrown upwards follows a path known as parabolic motion. This is because the path it takes resembles the shape of a parabola, which is a symmetric curve. The parabolic motion occurs when an object is under the influence of gravity, and it describes both the upward and downward arcs of the trajectory.
This shape results because the vertical acceleration due to gravity is constant, affecting the ball’s velocity in a uniform manner. The equation used to describe this motion is quadratic, typically written as:

• \( h = -16t^2 + v_0 t \)

Here, \( h \) represents the height of the ball at time \( t \), \( v_0 \) is the initial velocity, and the term \(-16t^2\) represents the effect of gravity in feet per second squared.

Vertex of Parabola

In the context of parabolic motion, the vertex of the parabola represents the highest point reached by the object in motion. For our exercise, this highest point signifies the greatest height the ball reaches when thrown upwards. Mathematically, the vertex can be calculated using the formula:

• \( t = \frac{-b}{2a} \)

Applying this formula to our equation \(-16t^2 + 40t\), we derive the time \( t \) at which the maximum height occurs. In this specific scenario:

• \( a = -16 \)
• \( b = 40 \)
• \( t = \frac{-40}{2(-16)} = 1.25 \) seconds

At this time, the ball reaches its highest altitude. To find the actual maximum height, substitute \( t = 1.25 \) back into the original height equation, yielding a maximum height of 25 feet.

Solving Quadratic Equations

Quadratic equations are essential in analyzing parabolic motion problems. These equations take the form

• \( ax^2 + bx + c = 0 \)

To solve these, the quadratic formula is an invaluable tool:

• \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

For example, when determining when the ball reaches heights of 24 feet or 48 feet, you plug in the values into the equation

• \( -16t^2 + 40t - h = 0 \)

Given specific values of the coefficients \(a\), \(b\), and \(c\), solve for \(t\) to find the times at which these heights are achieved. It’s crucial to rearrange the equation so that it equals zero before employing the quadratic formula.

Projectile Motion

Projectile motion encompasses the path taken by objects like balls when thrown, propelled, or dropped with an initial force. This exercise highlights a simple form of projectile motion, where the ball is thrown vertically. Here, the only force affecting motion after the throw is gravity, pulling the ball downward.
During projectile motion, several key points are of interest:

• Vertical launch speed \( v_0 \), which determines how high the ball will go.
• Time to reach certain heights or the ground, calculated by solving the quadratic equations.
• The highest point or vertex of the parabolic path.

After reaching its peak, the ball descends back to the ground, often analyzed by determining when \( h \) becomes zero. This sequence covers the complete arc from launch to landing.