Question

An object thrown directly upward is at a height of $h$ feet after $t$ seconds, where $h=-16{(t-3)}^2+150$. At what height, in feet, is the object $2$ seconds after it reaches its maximum height?

• A. $6$
• B. $86$
• C. $134$
• D. $150$
• E. $166$

Answer 1

The correct option is B $86$

Given,

$h=-16{(t-3)}^2+150$ (where, h is the height and t is time)

For maximum height

$\frac{dh}{dt}=-16\times 2(t-3)\Rightarrow -32(t-3)=0$

$t=3$

$\frac{d^{2}h}{d{t}^2}=-32$ (negative)

Therefore at $t=3sec$, $h$ will be maximum

Thus $2$ second after reaching maximum height$=3+2\Rightarrow 5sec$

$h=-16{(5-3)}^2+150$

$=-16\times 4+150$

$=150-64$

$h=86ft$