An object is thrown upwards

Question

An object is thrown upward at a speed of 116 feet per second by a machine from a height of 9 feet off the ground. The height  of the object after  seconds can be found using the equation When will the height be 117 feet? When will the object reach the ground?

I need this explained step by step, please. the video my teacher gave me does not help

Answer 1

just set the h(t) equation = 117 and solve for t, with the quadratic formula, t=-b/2a + or - (1/21)sqr(b^2-4ac)

117 = -16t^2 + 116t + 9

16t^2 -116t + 108 = 0

4t^2 - 29t + 27 = 0; a=4, b=-29, c=27, plug into the quadratic formula

t= 29/8 + or - (1/8)sqr(29^2 -16(27))

t = 3.625 + or - 2.619 = 6.244 or 1.006 seconds

It reaches 117 feet in about 1 second and at about 6 1/4 seconds

to find when it hits the ground, just set the h(t) function = 0 and solve for t

0= -16t^2 + 116t + 9

16t^2 - 116t -9 = 0

t = 116/32 + or - (1/32)sqr(116^2 + 36(16))

t = 3.625 + or - (1/32)sqr14032

t= 3.625 + or - 3.702 = 7.327 or a negative number you can ignore

It goes up and reaches 117 feet in one second. It continues to go up and then falls back to a height of 117 again on the way down after 6 1/4 seconds, then in a little over another second it hits the ground at t=7.327 seconds, or about 7 1/3 seconds.