Problem 14 - Basic Engineering Mathematics

Question

The height \(s\) metres of a mass projected vertically upwards at time \(t\) seconds is \(s=u t-\frac{1}{2} g t^{2}\). Determine how long the mass will take after being projected to reach a height of \(16 \mathrm{~m}\) (a) on the ascent and (b) on the descent, when \(u=30 \mathrm{~m} / \mathrm{s}\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

Step-by-Step Solution

Answer

It takes roughly 0.59 seconds on ascent and 5.52 seconds on descent to reach 16 m.

Step 1: Identify the equation and known values

The given equation is the height function for a mass projected upwards: \[ s = ut - \frac{1}{2} gt^2 \]where \( u = 30 \, \text{m/s} \), \( g = 9.81 \, \text{m/s}^2 \), and we are looking for \( t \) when \( s = 16 \, \text{m} \).

Step 2: Substitute known values into the equation

Substitute \( u = 30 \, \text{m/s} \), \( g = 9.81 \, \text{m/s}^2 \), and \( s = 16 \, \text{m} \) into the equation:\[ 16 = 30t - \frac{1}{2} \times 9.81 \times t^2 \] This simplifies to:\[ 16 = 30t - 4.905t^2 \]

Step 3: Rearrange the equation into a standard quadratic form

Rearrange the equation:\[ 4.905t^2 - 30t + 16 = 0 \]This is a standard quadratic equation of the form \( at^2 + bt + c = 0 \).

Step 4: Apply the quadratic formula

Use the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:- \( a = 4.905 \)- \( b = -30 \)- \( c = 16 \)Solve for \( t \):\[ t = \frac{-(-30) \pm \sqrt{(-30)^2 - 4 \cdot 4.905 \cdot 16}}{2 \cdot 4.905} \]\[ t = \frac{30 \pm \sqrt{900 - 313.92}}{9.81} \]\[ t = \frac{30 \pm \sqrt{586.08}}{9.81} \]

Step 5: Solve for the time values

Calculate \( \sqrt{586.08} \) which is approximately \(24.21\). Substitute back to find \( t \):\[ t = \frac{30 \pm 24.21}{9.81} \]Thus, \( t \) can be:\[ t_1 = \frac{30 + 24.21}{9.81} \approx 5.52 \, \text{seconds} \]\[ t_2 = \frac{30 - 24.21}{9.81} \approx 0.59 \, \text{seconds} \]

Step 6: Determine ascent and descent times

The first time \( t_1 \approx 0.59 \, \text{seconds}\) corresponds to the ascent, and the second time \( t_2 \approx 5.52 \, \text{seconds}\) corresponds to the descent.

Key Concepts

Quadratic Equation

The quadratic equation is a fundamental concept in mathematics, widely used in physics problems to describe various parabolic relationships. In the context of projectile motion, it allows us to determine the time at which a projectile will reach a certain height. The general form of a quadratic equation is:

• \( ax^2 + bx + c = 0 \)

Here, the variables \( a \), \( b \), and \( c \) are coefficients that define the particular parabola.
The quadratic formula is used to find the solutions for \( x \), which in our case translates to time \( t \) that satisfies this equation. The formula is:

• \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

With two potential solutions as indicated by the "\( \pm \)" sign, it reflects the dual nature of parabolic paths in projectile motion, where an object can reach a height both on the way up and the way down. Our specific scenario in this exercise uses values:

• \( a = 4.905 \)
• \( b = -30 \)
• \( c = 16 \)

These values are substituted into the quadratic formula to solve for \( t \). Understanding this process highlights the quadratic equation as a tool for solving real-world physical problems.

Kinematics

Kinematics is the branch of physics that studies the motion of bodies without considering the forces that cause them. In projectile motion problems, kinematics provides a framework for analyzing various parameters, such as displacement, velocity, and acceleration.
In this problem, we deal primarily with vertical motion, where the height \( s \) of a projectile is described by the equation:

• \( s = ut - \frac{1}{2} gt^2 \)

Here:

• \( u \) is the initial velocity, which is the velocity at which the object is launched.
• \( g \) is the acceleration due to gravity, acting downwards.
• \( t \) is the time variable, representing how long the object has been in motion.

This equation combines linear and quadratic terms to model how the height changes over time. It's crucial to understand that at its peak, the projectile's velocity is zero before it starts descending, which is a key consideration for determining ascent and descent times.

Physics Problem Solving

Physics problem solving is a structured approach to addressing and understanding physical phenomena. It involves various strategies that facilitate the comprehension of underlying physical laws through real-world applications.
The first step in solving physics problems is identifying the relevant equations and known values. Then, substituting these values accurately into the equations is crucial, which forms the base of analysis. Rearranging equations to simplify calculations, such as converting into standard forms, aids in reducing complexity.
Next, selecting appropriate mathematical methods, like using the quadratic formula, helps resolve parameters otherwise difficult to evaluate. Ensuring units are consistent and interpreting solutions correctly (as with ascent and descent in projectile motion) reflect real-world conditions.

• Identify known quantities.
• Substitute into the appropriate physical formula.
• Use mathematical tools to solve equations.
• Interpreting results to align with real-world scenarios.

This methodology enhances problem-solving skills, making them transferable across many types of physics problems beyond projectile motion.