Problem 91 - Precalculus

Question

Involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will eventually reach a maximum height and then fall to the ground. The equation that relates the height \(h\) of a projectile \(t\) seconds after it is projected upward is given by $$h=\frac{1}{2} a t^{2}+v_{0} t+h_{0}$$ where \(a\) is the acceleration due to gravity, \(h_{0}\) is the initial height of the object at time \(t=0,\) and \(v_{0}\) is the initial velocity of the object at time \(t=0 .\) Note that a projectile follows the path of a parabola opening down, so \(a<0\). An object is thrown upward, and the table below depicts the height of the ball \(t\) seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. $$\begin{array}{|c|c|} \hline t \text { (seconos) } & \text { HeiGHT (FEET) } \\ \hline 1 & 34 \\ \hline 2 & 36 \\ \hline 3 & 6 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Initial height \( h_0 = 16 \), initial velocity \( v_0 = 26 \), and acceleration \( a = -16 \).

Step 1: Understand the Parabolic Equation

Recognize that the height equation is given by \( h = \frac{1}{2} a t^2 + v_0 t + h_0 \), where \( a \) is the acceleration due to gravity (negative), \( v_0 \) is the initial velocity, and \( h_0 \) is the initial height. The data provided consists of time \( t \) and corresponding height \( h \) values.

Step 2: Set Up Equations Using Given Data

Using the time-height pairs: at \( t = 1 \), \( h = 34 \); at \( t = 2 \), \( h = 36 \); and at \( t = 3 \), \( h = 6 \). Substitute these values into the parabolic equation to produce three equations:1. \( 34 = \frac{1}{2} a (1)^2 + v_0 (1) + h_0 \)2. \( 36 = \frac{1}{2} a (2)^2 + v_0 (2) + h_0 \)3. \( 6 = \frac{1}{2} a (3)^2 + v_0 (3) + h_0 \)

Step 3: Simplify and Write the System of Equations

Transform each equation:1. \( 34 = \frac{1}{2} a + v_0 + h_0 \)2. \( 36 = 2a + 2v_0 + h_0 \)3. \( 6 = \frac{9}{2} a + 3v_0 + h_0 \) These are your system of equations based on the data.

Step 4: Eliminate h0 Using Equations 1 and 2

Subtract Equation 1 from Equation 2 to eliminate \( h_0 \):\( 36 - 34 = (2a + 2v_0 + h_0) - (\frac{1}{2} a + v_0 + h_0) \)This simplifies to:\( 2 = \frac{3}{2}a + v_0 \) (Equation 4)

Step 5: Eliminate h0 Using Equations 2 and 3

Subtract Equation 2 from Equation 3 to eliminate \( h_0 \):\( 6 - 36 = (\frac{9}{2} a + 3v_0 + h_0) - (2a + 2v_0 + h_0) \)This simplifies to:\( -30 = \frac{5}{2}a + v_0 \) (Equation 5)

Step 6: Solve for v0 and a

Use Equations 4 and 5 to solve for \( a \) and \( v_0 \):Subtract Equation 4 from Equation 5:\(-30 - 2 = (\frac{5}{2}a + v_0) - (\frac{3}{2}a + v_0) \)This gives:\(-32 = a \), so \( a = -16 \). Substitute \( a \) back into Equation 4:\( 2 = \frac{3}{2}(-16) + v_0 \)\( 2 = -24 + v_0 \)Solve for \( v_0 \):\( v_0 = 26 \)

Step 7: Solve for Initial Height h0

Use one of the original equations (Equation 1, 2, or 3) to find \( h_0 \). Using Equation 1:\( 34 = \frac{1}{2}(-16) + 26 + h_0 \)\( 34 = -8 + 26 + h_0 \)\( 34 = 18 + h_0 \)Solve for \( h_0 \):\( h_0 = 16 \)

Key Concepts

Parabolic Motion

In physics, when an object is thrown into the air, it follows a specific path called a parabola. This type of motion is referred to as parabolic motion. Parabolic motion describes the trajectory of any object under the influence of gravity alone, after initial propulsion. This path is characteristic of a projectile because gravity is the only force acting on it horizontally once the object is in motion.

The shape of the path, or the parabola, is affected by several factors, including the initial velocity and the angle at which the projectile is launched. This type of motion assumes that air resistance is negligible, which simplifies the calculations.

One important aspect of parabolic motion is the vertex of the parabola, which represents the highest point an object reaches during its trajectory. After the object reaches this highest point, it starts descending back towards the ground due to gravity. This concept is crucial for understanding how projectiles move in our problem.

Initial Velocity

The initial velocity (\( v_0 \) ) of a projectile is a vector quantity which describes the speed and direction at which the object is launched. In the equation \( h = \frac{1}{2} a t^2 + v_0 t + h_0 \), the term \( v_0 \) represents the velocity at time \( t=0 \).

The initial velocity is a critical component in determining how far and high a projectile will travel. A higher initial velocity means the projectile will typically travel further and higher, assuming all other factors remain constant.

Initial velocity is initially directed at an angle, typically above the horizontal axis. For upward motion, this means it has both horizontal and vertical components. In the context of our equation-solving problem, the value of \( v_0 \) was determined by solving the system of equations derived from the height data points.

Acceleration Due to Gravity

Acceleration due to gravity (\( a \)) is a key factor in projectile motion, acting downward on the projectile. This constant is denoted by \( g \) and typically approximated to a value of \(-9.8 m/s^2\), although in our exercise, the value was found to be \(-16\) (assuming feet).

This acceleration is negative because it opposes the initial motion of the projectile, pulling it back towards the earth regardless of the initial direction of motion. Gravity ensures that after reaching the maximum height, the projectile accelerates downward back to the ground.

Understanding this constant is crucial because it allows us to predict and calculate the movement of any object in free fall within a given gravitational field, excluding air resistance.

Equation Solving

Equation solving is the process of finding the values of the unknown variables in a mathematical problem. Here, we used systems of equations to determine the unknowns: initial height (\( h_0 \)), initial velocity (\( v_0 \)), and acceleration due to gravity (\( a \)).

The procedure involved setting up a system of equations based on known values from a height-time table. Each data point substituted into the motion equation yielded a unique equation. By manipulating this system, we canceled out variables and solved for the unknowns systematically.

• First, we substituted given heights and times into the parabolic motion equation to create individual equations.
• Next, we used algebraic manipulation to eliminate \( h_0 \) and solve for \( v_0 \) and \( a \).
• Finally, by substituting back, we determined the initial height.

This methodical approach highlights the power of algebra and calculus in physics for solving real-world problems, grounding theoretical knowledge into practical scenario-solving.