The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = −16t^2 + vt + c, where c is the initial height of the object above the ground. A model rocket is shot vertically up from a height of 6 feet above the ground with an initial velocity of 22 feet per second. Will it reach a height of 10 feet? Identify the correct explanation for your answer.
+128566 v = 22 c = 6
So we have that
-16t^2 + 22t + 6 = 10
-16 t^2 + 22t - 4 = 0 multiply through by - 1
16t^2 - 22t + 4 = 0 divide through by 2
8t^2 - 11t + 2 = 0
Using the quadratic formula
t = 11 ± sqrt ( 11^2 - 4 * 8 * 2) 11 ± sqrt (57)
________________________ = _____________
2 * 8 16
It first reaches 10 ft at [ 11 - sqrt (57) ] / 16 ≈ .216 sec
