Chapter 9.2, Problem 17P

A disc is thrown into the air with an upward velocity of 20 ft/s. Its height $h$ in feet after $t$ seconds is given by the function $h=-16t^2+20t+6.$

What is the maximum height the disc reaches? How long will it take the disc to reach the maximum height? How long does it take the disc to be caught 3 feet off the ground?

To determine

To find: The maximum height the disc reaches and time taken the disc to reach the maximum height and time taken the disc to be caught $3$ feet off the ground.

Answer to Problem 17P

The maximum height the disc reaches $=12.25\text{ feet}$

time taken the disc to reach the maximum height $t=0.625\text{ sec}$ .

time taken the disc to be caught $3$ feet off the ground $t=0.755\text{ sec}$ .

Explanation of Solution

Given information: A disc is thrown into the air with upward velocity of $20\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ . Its heights h in feet after t seconds is given by the function $h=-16t^2+20t+6$ .

Calculation:

A disc is thrown into the air with upward velocity of $20\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ . Its heights h in feet after t seconds is given by the function $h=-16t^2+20t+6$ .

Graph of function $h=-16t^2+20t+6$

According to the given graph

The maximum height the disc reaches $=12.25\text{ feet}$

and time taken the disc to reach the maximum height.

The maximum height the disc reaches $=12.25\text{ feet}$ .

function $h=-16t^2+20t+6$

Here, $h=12.25\text{ feet}$

  $\begin{array}{l}h=12.25\text{ feet}\\ 12.25=-16t^2+20t+6\\ 16t^2-20t+6.25=0\\ t=0.625\text{ sec}\end{array}$

time taken the disc to reach the maximum height $t=0.625\text{ sec}$ .

time taken the disc to be caught $3$ feet off the ground.

  $\begin{array}{l}h=3\text{ feet}\\ 3=-16t^2+20t+6\\ 16t^2-20t-3=0\\ t=1.38\text{ sec}\end{array}$

time taken the disc to be caught $3$ feet off the ground $t=(1.38-0.625)\text{ sec}$ .

time taken the disc to be caught $3$ feet off the ground $t=0.755\text{ sec}$ .