Chapter 1 - Equations and Inequalities - 1.5 Applications and Modeling with Quadratic Equations - 1.5 Exercises - Page 129: 7

Option B $$s=-16(2)^2+45(2)$$

The formula for the height in feet, $s$, of an object launched vertically at $45$ feet/sec after $t$ seconds is given as: $s=-16t^2+45t$ To find the height of the object after $2$ seconds, plug in $2$ for $t$: $$s=-16(2)^2+45(2)$$ which is option B