a person throws a stone into the air. The heigh, in feet h(t) after t secound is given by the following equation.h(t)=-16t2+64t+18a. what is the height of the stone after 3 secoundsb. when is the stone at a height of 36 feet?c. when does the stone reach the ground?

Answered: a. what is the height of the stone…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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a person throws a stone into the air. The heigh, in feet h(t) after t secound is given by the following equation.

h(t)=-16t²+64t+18

a. what is the height of the stone after 3 secounds

b. when is the stone at a height of 36 feet?

c. when does the stone reach the ground?

Expert Solution

Step 1

Consider the given height function,

$h (t) = - 16 t^{2} + 64 t + 18$

a) To find the height after 3 seconds, calculate $h (3)$ .

$\begin{array}{rcl} h (3) & = & - 16 {(3)}^{2} + 64 (3) + 18 \\ = & 66 \end{array}$

Hence, the height of the stone is 66 feet after 3 seconds.

b) To find the time when the height is 36 feet, set $h (t) = 36$ .

$\begin{array}{rcl} - 16 t^{2} + 64 t + 18 & = & 36 \\ - 16 t^{2} + 64 t - 18 & = & 0 \\ - 2 (8 t^{2} - 32 t + 9) & = & 0 \\ 8 t^{2} - 32 t + 9 & = & 0 \\ t & = & \frac{- (- 32) \pm \sqrt{{(- 32)}^{2} - 4 (8) (9)}}{2 (8)} \\ t & = & \frac{32 \pm \sqrt{736}}{16} \\ t & = & \frac{8 \pm \sqrt{46}}{4} \\ t & \approx & 3.7, 0.3 \end{array}$

Therefore, the height of the stone is 36 feet after 0.3 second and 3.7 seconds.

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