14. -/0.93 points details my notes sprecalc8 3.1.051. if a ball is thrown directly upward with a velocity of…
14. -/0.93 points details my notes sprecalc8 3.1.051. if a ball is thrown directly upward with a velocity of 20 ft/s, its height (in feet) after t seconds is given by y = 20t - 16t². what is the maximum height (in ft) attained by the ball? (round your answer to the nearest whole number.) ft need help? read it watch it submit answer
Answer
Explanation:
Step1: Identify the function type
The height - function $y = 20t-16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 20$, and $c = 0$.
Step2: Find the time at which 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 $b = 20$ into the formula, we have $t=-\frac{20}{2\times(-16)}=\frac{20}{32}=\frac{5}{8}$.
Step3: Find the maximum height
Substitute $t = \frac{5}{8}$ into the height - function $y = 20t-16t^{2}$. [ \begin{align*} y&=20\times\frac{5}{8}-16\times(\frac{5}{8})^{2}\ &=\frac{100}{8}-16\times\frac{25}{64}\ &=\frac{100}{8}-\frac{25}{4}\ &=\frac{100 - 50}{8}\ &=\frac{50}{8}\ &=\frac{25}{4}=6.25\approx6 \end{align*} ]
Answer:
6