Chapter 7, Problem 11STP

To determine

To Find :

The maximum height the object reaches.

Answer to Problem 11STP

  $\begin{array}{l}\text{The missile reaches its maximum height in 1}\text{.25 seconds and the maximum }\\ \text{height is 95 feet}\text{.}\end{array}$

Explanation of Solution

Given :

Suppose a projectile is launched into the air from a platform.The formula $h=-16t^2+40t+70$ relates the height h of the object (in feet) and the time t since it was launched (in seconds).

Concept Used :

The Standard form of a Quadratic Equation is :

  $y=a{x}^2+bx+c$

If a < 0 :

The parabola that represents the graph of the equation open downwards and the vertex is the maximum point of the parabola.

  $\begin{array}{l}\text{x-value of the vertex is given by:}\\ x=-\frac{b}{2a}\end{array}$

Calculation:

  $\begin{array}{l}Given:\\ h=-16t^2+40t+70\\ Comparing\text{ it with the standard form of quadratic equation:}\\ a=-16<0,b=40,c=70\\ \text{Since a < 0 , the parabola open downwards and the vertex is }\\ \text{the maximum point of the parabola}\text{.}\\ \text{t-value of the vertex is :}\\ t=-\frac{b}{2a}=-\frac{40}{2(-16)}=\frac{5}{4}\\ \text{Substitute the value of t in the given equation to find the maximum height:}\\ h\left(\frac{5}{4}\right)=-16{\left(\frac{5}{4}\right)}^2+40\left(\frac{5}{4}\right)+70=95\\ So,\text{ the missile reaches its maximum height in 1}\text{.25 seconds and the maximum }\\ \text{height is 95 feet}\text{.}\end{array}$