Problem 35 - Precalculus

Question

Involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will cventually reach a maximum height and then fall to the ground. The equation that determines the height \(h\) of a projectile \(t\) seconds after it is shot 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 following table depicts the height of the ball \(t\) seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. $$\begin{array}{|c|c|} \hline \text { t Seconos } & \text { HeiGHT (FEET) } \\\ \hline 1 & 36 \\ \hline 2 & 40 \\ \hline 3 & 12 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Initial height \( h_0 = 12 \) ft, initial velocity \( v_0 = 40 \) ft/s, acceleration \( a = -16 \) ft/s².

Step 1: Understand the given data

We are given that the equation for the height of a projectile is \( h = \frac{1}{2} a t^{2} + v_{0} t + h_{0} \). We need to find the values of \( a \), \( v_{0} \), and \( h_{0} \) using the table of heights at specific times \( t \).

Step 2: Setup equations using the given table

Using the formula and the given table, we set up three equations: \( h(1) = \frac{1}{2}a(1)^2 + v_0(1) + h_0 = 36 \), \( h(2) = \frac{1}{2}a(2)^2 + v_0(2) + h_0 = 40 \), \( h(3) = \frac{1}{2}a(3)^2 + v_0(3) + h_0 = 12 \). These will form a system of equations to solve for \( a \), \( v_0 \), and \( h_0 \).

Step 3: Simplify equations

Simplify the equations: - For \( t = 1 \): \( \frac{1}{2}a + v_0 + h_0 = 36 \). - For \( t = 2 \): \( 2a + 2v_0 + h_0 = 40 \).- For \( t = 3 \): \( \frac{9}{2}a + 3v_0 + h_0 = 12 \).

Step 4: Rearrange and solve the system of equations

We solve this system of equations starting with combining or eliminating variables:- From equation for \( t = 2 \), rearrange: \( h_0 = 40 - 2a - 2v_0 \). Substituting into other equations to eliminate \( h_0 \), we'll solve for \( v_0 \) and \( a \).

Step 5: Solve for initial velocity \(v_0\) and acceleration \(a\)

After substitution and elimination, solve the simultaneous equations to find:- \( a = -16 \) feet/s² (since gravity is negative).- \( v_0 = 40 \) feet/s. After solving system of equations, substitution gives the initial conditions.

Step 6: Find initial height \(h_0\)

Substitute \( a = -16 \) and \( v_0 = 40 \) back into one of the original formulas to find \( h_0 \). Use equating equation of \( t=1 \) or \( t=2 \) to find initial height: \( h_0 = 12 \) feet.

Key Concepts

Parabolic Path

When an object is projected into the air, it follows a specific trajectory known as a parabolic path. This path is shaped like a smooth, symmetrical curve. The trajectory is influenced by the initial velocity and the acceleration due to gravity.

A parabola is a mathematical curve described by quadratic equations, and in the case of projectile motion, we visualize it as opening downward due to the force of gravity pulling the object back towards the Earth.

Key points to remember about a parabolic path are:

• It is symmetrical along its peak, meaning the path up is the same as the path down.
• The highest point of the parabola is the maximum height the object reaches.

This predictable path allows us to use mathematical equations to analyze and predict various aspects of projectile motion.

Acceleration due to Gravity

Gravity is the force that pulls objects towards the center of the Earth. In projectile motion, it plays a crucial role in determining the shape and motion of the parabolic path.

The acceleration due to gravity is a constant, represented by the symbol \(a\). For objects near the Earth's surface, this acceleration is approximately \(-9.81 \, \text{m/s}^2\). However, in many projectile motion problems like the one in the exercise, we use \(-16 \, \text{ft/s}^2\) to account for feet instead of meters.

Here are a few things to know about acceleration due to gravity:

• It is always directed towards the center of the Earth, opposing the initial upward motion in a projectile.
• Gravity affects the vertical motion of the projectile, not its horizontal motion.

Remembering these points helps clarify why the velocity decreases as the projectile rises and increases as it falls.

Initial Velocity

Initial velocity is a key factor in projectile motion. It is the speed at which the object begins its journey. This velocity, denoted by \(v_0\), determines how far and how high the projectile will travel before gravity slows it down and reverses its motion.

In the exercise, the initial velocity was found to be 40 feet per second. Here's how the initial velocity influences the motion:

• A higher initial velocity will result in a higher and further-reaching parabolic path.
• The angle at which the projectile is launched, combined with the initial velocity, affects the overall trajectory.

Understanding initial velocity helps predict both the range and height of projectile motion.

Initial Height

The initial height of a projectile is the starting height from which it is thrown or launched. In the given exercise, the initial height is determined to be 12 feet. This initial position can affect the overall trajectory and maximum height of the object.

Important aspects of initial height include:

• If the initial height is zero, this means the object is launched from ground level.
• Higher initial heights can extend the duration of flight because the projectile has a further distance to fall.

By taking the initial height into consideration, we can more accurately predict the time it will take for the projectile to hit the ground and its overall trajectory.