Question

6.2 seconds The height, h, in feet of an object above the ground is given by h = −16t2 + 64t + 160, where t is the time in seconds. How long does it take the object to hit the ground? (to the nearest tenth of a second)

A)
5.1 seconds


B)
5.3 seconds


C)
5.7 seconds


D)
6.2 seconds

Answer 1

ANSWER

C) 5.7 seconds

EXPLANATION

The height of the object is given by:

$h(t) = - 16 {t}^{2} + 64t + 160$

If the object hit the ground, then the height is zero.

$- 16 {t}^{2} + 64t + 160 = 0$

Divide through by -16

${t}^{2} - 4t - 10 = 0$

Where a=1, b=-4 and c=-10

We substitute into the quadratic formula to obtain,

$t= \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$t= \frac{ - - 4\pm \sqrt{ {( - 4)}^{2} - 4( 1)( - 10)} }{2(1)}$

$t= \frac{ 4\pm \sqrt{56} }{2}$

$t= \frac{ 4\pm 2\sqrt{14} }{2}$

t=2-√14 or t=2+√14

Time cannot be negative.

Hence, t=5.7 seconds to the nearest tenth.