Answer to Question #192028 in Algebra for Katie

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

We have given the equation of height,

$h(t) = -16t^2+56t+25$

$h'(t) = -32t+56$

For maximum height,

$h'(t) = 0$

Hence,

$-32t+56 = 0$

$t = \dfrac{56}{32} = 1.75s$

at $t = 1.75s$

Maximum Height $= -16(1.75)^2+56(1.75)+25$

$= -28+98+25$

$= 95feet$