Problem 100 - Elementary Algebra

Question

The height in feet of an object dropped from a 20-foot platform is given by \(h(t)=-16 t 2+20,\) where \(t\) represents the time in seconds after the object has been dropped. How long does it take the object to hit the ground?

Step-by-Step Solution

Answer

The object takes \(\frac{\sqrt{5}}{2}\) seconds to hit the ground.

Step 1: Set the equation equal to 0

The object hits the ground when its height is 0. Therefore, we need to set the height equation equal to zero: \[-16t^2 + 20 = 0\]

Step 2: Solve for t using algebraic steps

First, subtract 20 from both sides to isolate the quadratic term:\[-16t^2 = -20\]Next, divide both sides by \(-16\) to solve for \(t^2\):\[t^2 = \frac{20}{16}\]Simplify the fraction:\[t^2 = \frac{5}{4}\]

Step 3: Solve for t using square roots

Take the square root of both sides to solve for \(t\):\[t = \pm \sqrt{\frac{5}{4}}\]We only consider the positive root because time cannot be negative:\[t = \sqrt{\frac{5}{4}}\]

Step 4: Simplify the expression

Further simplify the square root:\[t = \frac{\sqrt{5}}{2}\]

Key Concepts

Projectile Motion

Projectile motion is a fascinating topic that deals with objects that are launched into the air and affected only by gravity. In these types of problems, we often use physics to analyze the path or trajectory of an object. Such objects could be balls, stones, or even rockets. When an object is dropped, as in the given exercise, it is a special type of projectile motion known as free fall.

Here, the object is released from rest, so its initial velocity is zero. The only force acting on it is gravity, which gives it a constant acceleration downwards of approximately \( -9.8 \, \text{m/s}^2 \) on Earth (though the equation here uses feet, specifying \( -16 \, \text{ft/s}^2 \) for calculations in feet).

This exercise gives us an equation for the height of an object: \( h(t) = -16t^2 + 20 \). The \( -16t^2 \) part accounts for the acceleration due to gravity, and the \( 20 \) represents the initial height of the platform. As time progresses, gravity pulls the object down and eventually it reaches the ground, which is when the height is zero.

Solving Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the unknown variable, usually represented as \(x\), is 2. These equations typically have the form \( ax^2 + bx + c = 0 \).

In the context of our problem, we encounter a quadratic because of the term \( -16t^2 \), which varies as the square of time, \(t\). Solving quadratic equations is crucial in determining when an object will hit the ground in projectile motion problems.

There are various methods for solving quadratic equations, including

• Factoring
• Using the quadratic formula
• Completing the square
• The square root method, which is particularly useful when the equation is in the form \( x^2 = k \)

In our solution, after simplifying the original problem statement by setting it to zero, we isolate \(t^2\) before using the square root method to solve it.

Square Root Method

The square root method is a straightforward way to solve quadratic equations particularly suitable for equations without a linear \(t\) term, i.e., those in the form \( t^2 = k \). This method is elegant for its simplicity in cases where quadratics can be rearranged into a perfect square.

In the exercise problem, after isolating the quadratic term and simplifying the equation to \( t^2 = \frac{5}{4} \), we see it's ready for applying the square root method. The next step is straightforward: take the square root of both sides.

Do remember:

• Taking the square root introduces both positive and negative roots: \( t = \pm\sqrt{\frac{5}{4}} \).
• Since time cannot be negative, we discard the negative root for \( t \).

Ultimately, we find \( t = \frac{\sqrt{5}}{2} \), which represents the time in seconds when the object hits the ground. This method is efficient in such clear-cut scenarios and is widely used in projectiles without initial velocity.