Question

if a projectile is fired straight upward from the ground with an initial speed of 32 feet per second, then its height h in feet after t seconds is given by the function h(t)= - 16t² + 32t. find the maximum height of the projectile. (simplify your answer.)

Explanation:

Step1: Identify the function

The height function of the projectile is $h(t)=- 16t^{2}+32t$. This is a quadratic - function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 32$, and $c = 0$.

Step2: Find the time at which the maximum occurs

For a quadratic function $y=ax^{2}+bx + c$, the $x$ - coordinate (in our case, the time $t$) of the vertex is given by $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 32$ into the formula, we have $t=-\frac{32}{2\times(-16)}=\frac{-32}{-32}=1$ second.

Step3: Find the maximum height

Substitute $t = 1$ into the height function $h(t)$. So, $h(1)=-16\times(1)^{2}+32\times1=-16 + 32=16$ feet.

Answer:

16 feet