Question 1029190
You throw the rocket up in the air with an initial velocity of 50 feet per second, and the rocket leaves your hand 6 feet above the ground. if you catch it when it falls back to a height of 5 feet, how long was the rocket in the air? does the increase in initial velocity increase or decrease the air time of the rocket? 
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You didn't spec acceleration due to gravity.
-32 ft/sec/sec is commonly used.
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h(t) = gt^2/2 + vt + h0 where g is gravity, v = 50 ft/sec, h0 = 6 feet
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h(t) = -16t^2 + 50t + 6 : h(t) = height in feet, t = seconds
Solve for t when h(t) = 5
h(t) = -16t^2 + 50t + 6 = 5
-16t^2 + 50t + 1 = 0
*[invoke solve_quadratic_equation -16,50,1]
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Ignore the negative solution
t =~ 3.145 seconds