if a ball is thrown directly upward with a velocity of 24 ft/s, its height (in feet) after t seconds is…
if a ball is thrown directly upward with a velocity of 24 ft/s, its height (in feet) after t seconds is given by y = 24t - 16t². what is the maximum height attained by the ball? (round your answer to the nearest whole number.) ft
Answer
Explanation:
Step1: Identify the function type
The height - function $y = 24t-16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 24$, and $c = 0$.
Step2: Find the time $t$ at which the maximum occurs
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 $b = 24$ into the formula, we have $t=-\frac{24}{2\times(-16)}=\frac{24}{32}=\frac{3}{4}$.
Step3: Find the maximum height
Substitute $t = \frac{3}{4}$ into the height - function $y = 24t-16t^{2}$. $y=24\times\frac{3}{4}-16\times(\frac{3}{4})^{2}$ $y = 18-16\times\frac{9}{16}$ $y=18 - 9$ $y = 9$.
Answer:
9