An object is thrown upward at a speed of 92 feet per second by a machine from a height of 18 feet off the ground.

Question

An object is thrown upward at a speed of 92 feet per second by a machine from a height of 18 feet off the ground. The height hh of the object after tt seconds can be found using the equation h=−16t^2+92t+18

When will the height be 56 feet?

When will the object reach the ground? 

Answer 1

h(t) = -16t^2 + 92t + 18

h= 56 = -16t^2 + 92t + 18

16t^2 -92t + 38 = 0

8t^2 -46t + 19 = 0

t = 46/16 + or - (1/16)sqr(46^2 -4(8)(19)) = about .45 seconds = about 1/2 a second to reach 56 feet

on the way up and about 5.3 seconds to reach 56 feet on the way down

h=0 = 16^2 -92t -18

8t^2 -46t - 9 = 0

t = 46/16 + (1/16)sqr(46^2 +4(8)(9)) = about 5.9 seconds to reach the ground