SOLUTION: 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

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?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
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

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=2564 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.0198736126465204, 3.14487361264652. Here's your graph:

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Ignore the negative solution
t =~ 3.145 seconds
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