Chapter 3.2, Problem 88PS

Solution

The simplified expression for $h_0=0$ .

  $d=\frac{{\left(v_0\right)}^2}{g}$

Given:

The maximum horizontal distance d that an object can travel when launched at an optimum angle of projection is given by $d=\frac{v_0\sqrt{{\left(v_0\right)}^2+2g{h}_0}}{g}$ where $h_0$ is the object’s initial height, $v_0$ is its initial speed, and g is the acceleration due to gravity.

Concept Used:

• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• Product of power property: $a^m\cdot a^n=a^{m+n}$
• Quotient of power property: $\frac{a^m}{a^n}=a^{m-n}$
• Product property of radicals: $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$
• Quotient property of radicals: $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• The exponent form of the nth root of a number a is given by $\sqrt[n]{a}=a^{\frac{1}{n}}$

Calculation:

In order to simplify the given expression, substitute $h_0=0$ into the given model and simplify further as shown below

  $\begin{array}{c}d=\frac{v_0\sqrt{{\left(v_0\right)}^2+2g{h}_0}}{g}\\ d=\frac{v_0\sqrt{{\left(v_0\right)}^2+2g\cdot 0}}{g}\\ d=\frac{v_0\sqrt{{\left(v_0\right)}^2+0}}{g}\\ d=\frac{v_0\sqrt{{\left(v_0\right)}^2}}{g}\\ d=\frac{v_0\cdot v_0}{g}\\ d=\frac{{\left(v_0\right)}^2}{g}\end{array}$

Thus, the simplified expression is $d=\frac{{\left(v_0\right)}^2}{g}$ .