Question

An object is thrown upward at a speed of 194 feet per second by a machine from a height of 6 feet off the ground. The height hh of the object after tt seconds can be found using the equation h(t)=−16t2+194t+6h(t)=-16t2+194t+6

When will the height be 522 feet?
(Write both results, using a comma between). Round answers to the nearest hundredth.

When will the object reach the ground?

Answer 1

It will at height 522 at 3.94 , 8.18   

object reach ground at 12.16

given equation for height h = -16t? 1946 +6 When it reach a height 522. hé) =522. apply this to equahon and find -16 +2 +192 +6=522 on rearranging this -16 +241941 - 516=0 now solve this quadratic equation. we can use quadratic quadratic formula by quadratic, formula, solution for ax²+bx+c=0 x = -b ± √b2_490 29 for our equshon a=-16 b=194, (=516. so to = -194 ± √1942_4X1-16-516 T2x-16 t = -194 +√4612 -194 + 67.9117 -32 - 32 t = -194+67.9117 -32 =3.94 -194+6791178.18 OY TA -32

so answer ave 3.94, 8.18 h(t) = 0 when object reach ground apply this and Pind t -16 +2+194t +6=0 Use quadratic formula to solve this t= -194 I/194²_4+(16) x6 2x-16 te -194 + 194.98717 - 32 te -194.910 +194.98717 = -0.03 -32 07 = 12.16 so t = -194 +94-98717 - 32 object reach ground at t=12.16