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
Accepted Solution
A:
ANSWERC) 5.7 secondsEXPLANATIONThe height of the object is given by:[tex]h(t) = - 16 {t}^{2} + 64t + 160[/tex]If the object hit the ground, then the height is zero.[tex]- 16 {t}^{2} + 64t + 160 = 0[/tex]Divide through by -16[tex] {t}^{2} - 4t - 10 = 0[/tex]Where a=1, b=-4 and c=-10We substitute into the quadratic formula to obtain,[tex]t= \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} [/tex][tex]t= \frac{ - - 4\pm \sqrt{ {( - 4)}^{2} - 4( 1)( - 10)} }{2(1)} [/tex][tex]t= \frac{ 4\pm \sqrt{56} }{2} [/tex][tex]t= \frac{ 4\pm 2\sqrt{14} }{2} [/tex]t=2-√14 or t=2+√14Time cannot be negative.Hence, t=5.7 seconds to the nearest tenth.