## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Review Exercises - Page 432: 92

#### Answer

Between 1 s and 2 s, the projectile's height exceeds $32$ feet.

#### Work Step by Step

We are given that $s(t)=-16t^2+v_0t +s_0$ Set $v_0=48$ Thus, $0=-16(0)^2+(48)(0) +s_0$ so, $s_0=0$ Now, $-16t^2+48t=-32 \implies t^2-3t+2=0$ This implies that $(t-2)(t-1)=0$ or, $t= 1 s, 2 s$ Therefore, between 1 s and 2 s, the projectile's height exceeds $32$ feet.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.