Problem 36 - 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. $$\begin{array}{|c|c|} \hline \text { t Seconos } & \text { Heiant (FEET) } \\\ \hline 1 & 84 \\ \hline 2 & 136 \\ \hline 3 & 156 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Initial height is 16 feet, initial velocity is 76 ft/s, and acceleration due to gravity is -16 ft/s².

Step 1: Understanding the Equation

The equation for the vertical motion of a projectile is given by \[ h = \frac{1}{2} a t^2 + v_0 t + h_0 \]where:- \(a\) is the acceleration due to gravity (negative because the projectile moves against gravity when going up),- \(v_0\) is the initial velocity,- \(h_0\) is the initial height, - \(t\) is the time in seconds,- \(h\) is the height at time \(t\).

Step 2: Extract Information from the Data

From the table provided:- When \(t=1\), \(h=84\).- When \(t=2\), \(h=136\).- When \(t=3\), \(h=156\).These points will help form a system of equations using the height equation.

Step 3: Form the System of Equations

Substitute the values of \(t\) and \(h\) into the equation for each time:- For \(t=1\), \(h=84\): \[\frac{1}{2} a (1)^2 + v_0 (1) + h_0 = 84\]- For \(t=2\), \(h=136\): \[\frac{1}{2} a (2)^2 + v_0 (2) + h_0 = 136\]- For \(t=3\), \(h=156\): \[\frac{1}{2} a (3)^2 + v_0 (3) + h_0 = 156\]These lead to three equations:1. \(\frac{1}{2} a + v_0 + h_0 = 84\)2. \(2a + 2v_0 + h_0 = 136\)3. \(\frac{9}{2} a + 3v_0 + h_0 = 156\)

Step 4: Solve the System of Equations

First, simplify and rearrange the equations:1. \(\frac{1}{2} a + v_0 + h_0 = 84\)2. \(2a + 2v_0 + h_0 = 136\)3. \(\frac{9}{2} a + 3v_0 + h_0 = 156\)To eliminate \(h_0\),- Subtract equation 1 from equation 2: \[ 2a + 2v_0 + h_0 - (\frac{1}{2} a + v_0 + h_0) = 136 - 84 \] \[ \frac{3}{2} a + v_0 = 52 \] - Subtract equation 2 from equation 3: \[ \frac{9}{2} a + 3v_0 + h_0 - (2a + 2v_0 + h_0) = 156 - 136 \] \[ \frac{5}{2} a + v_0 = 20 \] Now solve these two new equations:1. \(\frac{3}{2} a + v_0 = 52\)2. \(\frac{5}{2} a + v_0 = 20\)Subtract equation 1 from equation 2:\[\left(\frac{5}{2} a + v_0\right) - \left(\frac{3}{2} a + v_0\right) = 20 - 52\]\[\frac{2}{2} a = -32 \ a = -16 \] Then substitute \(a = -16\) back into \( \frac{3}{2} a + v_0 = 52 \): \[ \frac{3}{2}(-16) + v_0 = 52 ewline -24 + v_0 = 52 ewline v_0 = 76 \]}},{

Step 5: Find the Initial Height

Use the value of \(a\) and \(v_0\) in one of the original equations to find \(h_0\), for example in equation 1:\[\frac{1}{2}(-16) + 76 + h_0 = 84\]\[-8 + 76 + h_0 = 84\]\[68 + h_0 = 84\]\[h_0 = 16\]

Key Concepts

Vertical Motion

In projectile motion, vertical motion plays a crucial role as it is dominated by the effects of gravity. When an object is thrown upwards, it's only a matter of time before it begins to fall back down due to gravity's pull. This vertical journey of the projectile follows a systematic path.

• Initially, the object rises upward, slowing down due to opposing gravity.
• It reaches a maximum height where its upward velocity becomes zero.
• Subsequently, it accelerates downward, increasing speed as it descends.

The height of the projectile over time can be determined using the equation:
\[ h = \frac{1}{2} a t^2 + v_0 t + h_0 \]Here, 'h' is the height, 't' is the time, 'a' represents acceleration due to gravity (a negative value since it's a downward force), 'v_0' is the initial velocity, and 'h_0' is the initial height.

Parabolic Trajectory

In the context of projectile motion, the path followed by the object is known as a parabolic trajectory. This trajectory is the result of two components:

• Horizontal motion, which remains uniform.
• Vertical motion, influenced by gravity.

Together, these components cause the projectile to trace out a curved path similar to the shape of a parabola. This characteristic downward-opening parabola results because gravity acts as a force pushing the projectile back to the ground. The equation that describes this motion is quadratic, indicating the signature parabolic shape:
\[ y = ax^2 + bx + c \]where the acceleration due to gravity 'a' affects how quickly the projectile falls back to earth. The initial height and velocity influence how high and far the object travels along this parabolic path.

System of Equations

Solving problems in projectile motion often requires setting up a system of equations. This allows us to find unknown variables like the initial height, initial velocity, and the acceleration due to gravity:

• Use different points in time to substitute into the height equation.
• Formulate multiple equations for different times 't'.

For example, using heights at different times, we derive equations:
\( \frac{1}{2}a + v_0 + h_0 = h_1 \)
\( 2a + 2v_0 + h_0 = h_2 \)
\( \frac{9}{2}a + 3v_0 + h_0 = h_3 \)
By having such a system, we are equipped to solve for the unknowns using algebraic methods like substitution or elimination. This step-by-step solving process is fundamental in understanding and applying the principles of physics in projectile motion.

Acceleration Due to Gravity

Acceleration due to gravity, commonly denoted by 'g', is the constant acceleration that pulls objects back towards Earth. On Earth, this value is approximately -9.8 m/s² (or -32 ft/s² when working in feet), and it is crucial in all projectile motion calculations.

• It causes the upward motion of projectiles to slow until they reach a peak and then accelerates them back downward.
• In equations, 'a' is the acceleration due to gravity and is negative due to the downward pull.

Being a constant, it simplifies calculations as it remains unchanged irrespective of the time or conditions. Its role in the projectile motion equation is central. It ensures that the overall motion is realistic and adheres to the physical laws governing our universe. Understanding its influence helps predict both the flight and the landing of projectiles accurately.