The height h in feet of an object after t seconds is given by the fraction h = -16t^2 + 30t + 8. How long will it take the object to hit the ground? Round answer to the nearest thousandth?

Question

2733 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-06T06:30:53

It will take $2.112$ $2.112$ seconds for object to hit ground.

The height of ground level is considered as $0$ $0$ .

as $h = - 16 t^{2} + 30 t + 8$ $h=-16t^2+30t+8$ , it will be zero, when

$- 16 t^{2} + 30 t + 8 = 0$ $-16t^2+30t+8=0$

or $16 t^{2} - 30 t - 8 = 0$ $16t^2-30t-8=0$ anddividing by $2$ $2$

$8 t^{2} - 15 t - 4 = 0$ $8t^2-15t-4=0$

Using quadratic formula $t = \frac{- (- 15) \pm \sqrt{{(- 15)}^{2} - 4 \times 8 \times (- 4)}}{16}$ $t=(-(-15)+-sqrt((-15)^2-4xx8xx(-4)))/16$

= $\frac{15 \pm \sqrt{225 + 128}}{16}$ $(15+-sqrt(225+128))/16$

= $\frac{15 \pm \sqrt{353}}{16}$ $(15+-sqrt353)/16$

= $\frac{15 \pm 18.7883}{16}$ $(15+-18.7883)/16$ , but as we cannot have $t$ $t$ negative

$t = \frac{33.7883}{16} = 2.112$ $t=33.7883/16=2.112$ seconds