Question

a projectile is thrown upward so that its distance above the ground after t seconds is h = -16t^2 440t. after how many seconds does it reach its maximum height?

Answer 1

When the projectile reaches
its maximum height, the slope of the projectile is horizontal at that maximum point, hence its derivative is zero.

Given

$h(t) = - 16 {t}^{2} + 440t$

The derivative is given by:
$h'(t) = - 32 {t} + 440$

At maximum height,

$h'(t) = 0$

$\Rightarrow - 32 {t} + 440 =0$

Now let us solve for t, to obtain;

$\ - 32 {t} = - 440$

$\Rightarrow t = \frac{ - 440}{ - 32}$

$\Rightarrow t = 13.75s$

$Hence the projectile reached the maximum height after 13.75 seconds.$

Answer 2

Velocity=-32t+440
At maximum height, v=0
0=-32t+440
-440=-32t
t=13.75 seconds