Problem 92 - 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. $$\begin{array}{|c|c|} \hline t \text { (seconos) } & \text { Heiant (FEET) } \\ \hline 1 & 54 \\ \hline 2 & 66 \\ \hline 3 & 46 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Initial height: 10 ft, initial velocity: 60 ft/s, gravity: -32 ft/s².

Step 1: Understand the Problem

Given the equation \(h=\frac{1}{2} a t^{2}+v_{0} t+h_{0}\), we need to determine three unknowns: the initial height \(h_0\), the initial velocity \(v_0\), and the acceleration due to gravity \(a\). The table provides the heights at specific time points \(t=1, 2, 3\) seconds.

Step 2: Create System of Equations

Using the heights given at specific times, plug the values \((t=1, 54), (t=2, 66), (t=3, 46)\) into the height equation to create three equations. \(1:\ 54 = \frac{1}{2} a (1)^2 + v_0 (1) + h_0\) \(2:\ 66 = \frac{1}{2} a (2)^2 + v_0 (2) + h_0\) \(3:\ 46 = \frac{1}{2} a (3)^2 + v_0 (3) + h_0\)

Step 3: Simplify Equations

Simplify the equations obtained:\(54 = \frac{1}{2} a + v_0 + h_0\) \(66 = 2a + 2v_0 + h_0\) \(46 = \frac{9}{2} a + 3v_0 + h_0\)

Step 4: Solve the System of Equations

Subtract the first equation from the second and the second from the third to eliminate \(h_0\) and form two equations:1. \(66 - 54 = (2a - \frac{1}{2} a) + (2v_0 - v_0)\) produces Equation (i): \(12 = \frac{3}{2} a + v_0\)2. \(46 - 66 = (\frac{9}{2} a - 2a) + (3v_0 - 2v_0)\) produces Equation (ii): \(-20 = 2.5a + v_0\).Now solve these two linear equations.

Step 5: Solve for Acceleration and Velocity

From Equation (i): \(v_0 = 12 - \frac{3}{2}a\). Substitute \(v_0\) into Equation (ii): \(-20 = 2.5a + (12 - \frac{3}{2}a)\)\(-20 = 2.5a + 12 - \frac{3}{2}a\)Simplify: \(-32 = a\). Hence, \(a = -32\). Then substitute \(a\) back to find \(v_0:\) \(v_0 = 12 - \frac{3}{2}(-32) = 12 + 48 = 60\).

Step 6: Solve for Initial Height

Substitute \(a = -32\) and \(v_0 = 60\) back into the equation for \(t=1\):\(54 = \frac{1}{2}(-32) + 60 + h_0\)\(54 = -16 + 60 + h_0\)\(54 = 44 + h_0\)\(h_0 = 10\).

Step 7: Conclusion

The initial height \(h_0\) is 10 feet, the initial velocity \(v_0\) is 60 feet/second, and the acceleration due to gravity \(a\) is \(-32\) feet/second extsuperscript{2}.

Key Concepts

Quadratic Equations

Projectile motion often involves quadratic equations, which describe paths that projectiles follow. In our problem, the quadratic equation used is the form: \[ h = \frac{1}{2} a t^2 + v_0 t + h_0 \] This equation defines the height \( h \) of an object after \( t \) seconds as a function of the acceleration due to gravity \( a \), the initial velocity \( v_0 \), and the initial height \( h_0 \). Here, the quadratic term \( \frac{1}{2} a t^2 \) represents the effect of gravity, providing the parabolic path.In general, quadratic equations can be identified because the highest power of the variable is two. This means the graph of a projectile's motion is a parabola, illustrating how the object travels up to its maximum height before descending. Quadratic equations are usually solved to find unknown variables using methods like factoring, completing the square, or the quadratic formula.

Gravity

Gravity is a natural force that pulls objects toward the center of the Earth. In projectile motion, gravity significantly influences an object's trajectory by providing a constant downward acceleration. Its standard value is approximately \(-9.8\) meters per second squared on Earth, however in the English system, it is typically \(-32\) feet per second squared as found in our solution.The acceleration due to gravity is denoted by \( a \) in the equation. Because this force always acts downward, it is considered a negative value in these scenarios. This negative acceleration helps form the parabolic path of the projectile, pulling it back to the ground after reaching maximum height. Understanding gravity's role in projectile motion can help you predict how and when an object will fall back.

Initial Velocity

The initial velocity, denoted as \( v_0 \), is the speed and direction at which an object begins its motion. In our exercise, this variable is crucial in determining how the projectile travels before gravity fully alters its course.You can think of initial velocity as how hard and swiftly you launch the object.

• A higher initial velocity means the object will travel farther and reach a higher point before gravity pulls it back down.
• Different initial velocities can result in varying flight times and distances.

In this problem, we discovered that \( v_0 = 60 \) feet per second, permitting the ball to ascend before gravity pulls it down. Initial velocity works with gravity and initial height to chart the object's complete movement.

Initial Height

Initial height, symbolized as \( h_0 \), is the starting elevation of an object before being affected by gravity. Initial height indicates where the object is positioned at the onset of its path.

• If you start with a greater initial height, the object will stay in the air longer.
• This factor allows for more time before it reaches the ground, as gravity will take longer to completely act upon it.

In our specific projectile exercise, the initial height \( h_0 \) was found to be 10 feet. This value was determined by considering the known heights at given times and solving the system of equations. The initial height, combined with initial velocity and gravity, greatly influences the total flight path of the projectile.