Question

watch the video and then solve the problem given below. click here to watch the video. if a baseball is projected upward from ground - level with an initial velocity of 160 feet per second, then its height is a function of time, given by s=-16t^2 + 160t. what is the maximum height reached by the ball? the maximum height reached by the ball is feet.

Explanation:

Step1: Identify the function type

The height function $s = - 16t^{2}+160t$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 160$, $c = 0$.

Step2: Find the time at which maximum occurs

For a quadratic function $y=ax^{2}+bx + c$, the $x$-coordinate (in our case, the time $t$) of the vertex is given by $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 160$ into the formula, we get $t=-\frac{160}{2\times(-16)}=\frac{160}{32}=5$ seconds.

Step3: Find the maximum height

Substitute $t = 5$ into the height - function $s=-16t^{2}+160t$. So $s=-16\times5^{2}+160\times5=-16\times25 + 800=-400 + 800 = 400$ feet.

Answer:

400