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 = ( - 440)/( - 32)`


`\Rightarrow t = 13.75s`


`<b>Hence the projectile reached the maximum height after 13.75 seconds.</b>`

Answer 2

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