Question

The formula $h\left(t\right)=-16t^2+32t+80$ gives the height $h$ above ground, in feet, of an object thrown, at $t=0$, straight upward from the top of an $80$ feet building
(a) What is the highest point reached by the object
(b) How long does it take the object to reach its highest point?
(c) After how many seconds does the object hit the ground?
(d) For how many seconds is the height of the object higher than $90$ feet?

Answer 1

Given $h\left(t\right)=-16t^2+32t+80$

Highest point reached by object is max. $\left(h\right(t\left)\right)$

$h^{\prime }\left(t\right)=-32t+32=0$

$\therefore t=1$

$\therefore \therefore h\left(t\right)=h\left(max\right)=-16+32+80$

$=96ft$

$\therefore$Highest point reached$=96ft$

Time taken to reach highest point$=1$ sec

When object hit ground, height, $h\left(t\right)=0$

$-16t^2+32t+80=0$

$t^2-2t-5=0$

$t=\frac{2+\sqrt{4+20}}{2}(\because t>0)$

$t=(1+\sqrt{6})$

Object hit ground after $(\sqrt{6}+1)$ sec

Time taken to reach $90ft\Rightarrow h\left(t\right)=90$

$-16t^2+32t+80=90$

$-16t^2+32t-10=0$

$8t^2-16t+5=0$

$t=\frac{16\pm \sqrt{256-160}}{16}=\frac{16\pm 4\sqrt{6}}{16}$

Time for which height of object is higher than $90ft$ is $(\frac{16+4\sqrt{16}}{16}-\frac{16-4\sqrt{6}}{16})=\sqrt{\frac{3}{2}}$