Question

a ball was shot from a cannon into the air with an upward velocity of 40 ft/sec. the equation that gives the height (h) of the ball at any time (t) is shown below. find the maximum height attained by the ball. h(t) = -16t² + 40t + 1.5

Explanation:

Step1: Find time of max height

For quadratic $h(t)=at^2+bt+c$, time $t=-\frac{b}{2a}$.
Here $a=-16$, $b=40$, so:
$t = -\frac{40}{2\times(-16)} = \frac{40}{32} = 1.25$ seconds

Step2: Substitute t into height formula

Plug $t=1.25$ into $h(t)=-16t^2+40t+1.5$:
$h(1.25) = -16(1.25)^2 + 40(1.25) + 1.5$
Calculate each term:
$-16(1.5625) = -25$, $40(1.25)=50$
Combine terms: $-25 + 50 + 1.5 = 26.5$

Answer:

26.5 feet