Question

If an object is dropped from the height of 54 feet the function h=-16t2 +54 gives the height of the object after t seconds approximately how long will it take before the object hits the ground

Answer 1

Answer:

t= +/- $\frac{3\sqrt{6}}{4}$

Step-by-step explanation:

set h=0, 0=-16$t^{2}$+54, add the -16$t^{2}$ to the other side, 16$t^{2}$=54, divide the 16 off, $t^{2}$=$\frac{54}{16}$, square root both sides, t= +/- $\sqrt[]{\frac{54}{16} }$ , t= +/- $\sqrt{54}$/4, t= +/- $\frac{3\sqrt{6}}{4}$ , the object will hit the ground when t= +/- $\frac{3\sqrt{6}}{4}$