Question

A watermelon is launched from a 25-foot tall platform at a carnival. The watermelon's height h (in feet) at time t seconds can be modeled by the equation h = -16t^2 + 56t + 25. Find the maximum height of the watermelon

Answer 1

Answer:

The maximum height of the watermelon is 74 feet.

Step-by-step explanation:

The watermelon reaches its maximum height when its velocity is zero. Mathematically speaking, velocity is the first derivative of the function height, that is:

$v(t) = \frac{d}{dt} h(t)$ (1)

Where $v(t)$ is the velocity, in feet per second.

If we know that $h(t) = -16\cdot t^{2}+56\cdot t +25$ and $v(t) = 0\,\frac{ft}{s}$, then the time associated with maximum height is:

$v(t) = -32\cdot t +56$ (2)

$-32\cdot t + 56 = 0$

$t = 1.75\,s$

Now we evaluate the function height at time found in the previous step: ($t = 1.75\,s$)

$h(t) = -16\cdot t^{2}+56\cdot t +25$

$h(1.75) = 74\,ft$

The maximum height of the watermelon is 74 feet.