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Magazine Chapter 1: Problem 41
Solve the equation for the indicated variable. $$h=\frac{1}{2} g t^{2}+v_{0} t ; \quad \text { for } t$$
Short Answer
Expert verified
The solution for \( t \) is \( t = \frac{-v_0 + \sqrt{v_0^2 + 2gh}}{g} \).
Step by step solution
01
Recognize the Equation Type
This is a quadratic equation in terms of the variable \( t \) because of the \( t^2 \) term. The equation is in the form \( h = \frac{1}{2} g t^2 + v_0 t \), which is similar to the standard quadratic equation \( ax^2 + bx + c = 0 \).
02
Rearrange the Equation
Subtract \( h \) from both sides of the equation to set it to 0: \( \frac{1}{2} g t^2 + v_0 t - h = 0 \).
03
Identify Coefficients
Identify the coefficients for the quadratic formula. Here, \( a = \frac{1}{2}g \), \( b = v_0 \), and \( c = -h \).
04
Apply the Quadratic Formula
Use the quadratic formula, which is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), to solve for \( t \). Substitute \( a = \frac{1}{2}g \), \( b = v_0 \), and \( c = -h \).
05
Simplify the Expression
Substitute the known values into the formula: \[ t = \frac{-v_0 \pm \sqrt{v_0^2 - 4 \left( \frac{1}{2}g \right)(-h)}}{2 \left( \frac{1}{2}g \right)} \]. Simplify further to give: \[ t = \frac{-v_0 \pm \sqrt{v_0^2 + 2gh}}{g} \].
06
Determine the Viable Solution
The equation for \( t \) gives two solutions due to the \( \pm \) sign. In physical situations, time is usually positive, so typically only the positive solution is meaningful: \[ t = \frac{-v_0 + \sqrt{v_0^2 + 2gh}}{g} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving for a variable
In many mathematical problems, solving for a variable means isolating one variable on one side of an equation. In the given exercise, we aim to solve the equation for the variable \( t \). This involves rearranging terms and applying appropriate strategies to isolate \( t \) in the equation. To solve for \( t \) in the quadratic equation \( h = \frac{1}{2} g t^2 + v_0 t \), we perform several key steps:
- Recognize the type of equation: Identify that it is a quadratic equation due to the \( t^2 \) term.
- Rearrange the equation: Bring all terms to one side so that it can be set to zero, resembling the standard form \( ax^2 + bx + c = 0 \).
- Identify coefficients: Extract the coefficients \( a \), \( b \), and \( c \) that define the equation, which are critical for further steps.
Quadratic formula
The quadratic formula is a universal tool used to find the solutions of any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is expressed as:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula allows us to directly compute the values of \( t \) by substituting the values of the coefficients \( a \), \( b \), and \( c \) from our quadratic equation.
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula allows us to directly compute the values of \( t \) by substituting the values of the coefficients \( a \), \( b \), and \( c \) from our quadratic equation.
- Calculation of discriminant: The expression under the square root sign, \( b^2 - 4ac \), is called the discriminant, and it determines the nature of the solutions.
- Solving: Substitute the coefficients into the formula. In our example with \( a = \frac{1}{2}g \), \( b = v_0 \), and \( c = -h \), result in two potential solutions.
- Interpretation: These solutions characterize the values where the quadratic equation is satisfied.
Physics application
The equation in question isn't just a mathematical exercise; it has practical implications in physics. Specifically, it is related to motion under constant acceleration, a common topic in physics, particularly when analyzing projectile motion or free-fall situations.
- Understanding Variables: In physics, \( h \) can represent height, \( g \) the acceleration due to gravity, usually approximated as 9.81 m/s虏, and \( v_0 \) the initial velocity of the object.
- Significance: Solving for \( t \) using these parameters helps predict when an object, subject to gravity and initial velocity, reaches a certain height \( h \).
- Practical Use: Often used in determining the time an object takes to reach a certain point during its motion, which is crucial in fields like engineering and space sciences.
