a ball is thrown vertically upward. after t seconds, its height h (in feet) is given by the function…
a ball is thrown vertically upward. after t seconds, its height h (in feet) is given by the function h(t)=40t - 16t². what is the maximum height that the ball will reach? do not round your answer. height: feet
Answer
Explanation:
Step1: Identify the function type
The height function $h(t)=40t - 16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 40$, and $c = 0$.
Step2: Find the time $t$ at maximum - height
For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case $t$ - coordinate) of the vertex is given by $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 40$ into the formula, we have $t=-\frac{40}{2\times(-16)}=\frac{40}{32}=\frac{5}{4}$.
Step3: Find the maximum height
Substitute $t = \frac{5}{4}$ into the height function $h(t)$. $h(\frac{5}{4})=40\times\frac{5}{4}-16\times(\frac{5}{4})^{2}$. First, calculate $40\times\frac{5}{4}=50$. Second, calculate $16\times(\frac{5}{4})^{2}=16\times\frac{25}{16}=25$. Then $h(\frac{5}{4})=50 - 25=25$.
Answer:
25