SOLUTION: hi, my question is following A toy rocket is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec. Using the function V(x)=-16t^2+vt

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: hi, my question is following A toy rocket is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec. Using the function V(x)=-16t^2+vt      Log On

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Question 283298: hi, my question is following
A toy rocket is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec. Using the function V(x)=-16t^2+vt+h, answer the following.
1. Give the function that describes the height of the rocket in terms of t?
2. Determine the time at which the rocket reaches its maximum height, and its maximum height in feet?
3. For what time interval will the rocket be more than 300 ft above the ground level?
4. for how many seconds will it hit the ground?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let x = t.

your equation becomes V(x) = -16x^2 + vt + h

h = 50 which is the height of the rocket at time point 0 (value of x is 0).

your equation becomes v(x) = -16x^2 + vx + 50

Since initial velocity is 200 ft/sec, then v = 200 and your equation becomes:

v(x) = -16x^2 + 200x + 50

maximum height should be when x = -b/2a

a = -16
b = 200
c = 50

maximum height should be when x = -200/-32 = 6.25

when x = 6.25, v(x) = 675 feet.

That's what the maximum height should be.

Set v(x) = 300 and solve for x.

300 = -16x^2 + 200x + 50

subtract 50 from both sides of this equation to get 0 = -16x^2 + 200x - 250.

solve for the roots of this equation by quadratic formula.

a = -16
b = 200
c = -250

(-b +/- sqrt(b^2 - 4ac))/(2a) =:
(-200 +/- sqrt(200^2 - 4*-16*-200))/(-32) =:
(-200 +/- sqrt(40000 - 12800))/(-32) =:
(-200 +/- sqrt(27200)))=32 =:
(-200 +/- 164.924225)/(-32)

x = 1.096117968 or x = 11.40388203

The rocket should be at or above 300 feet from x = 1.096117968 to x = 11.40388203

A graph of this equation is shown below:

graph%28400%2C600%2C-5%2C15%2C-100%2C1000%2C-16x%5E2+%2B+200x+%2B+50%2C300%2C675%29

A horizontal line was placed at y = 300 and 675 to show you the intersection points with the graph of the equation of the rocket.

At y = 675, the graph should be at the maximum point.

At 300, the graph should be at x = 1.09 and x = 11.4 roughly.

Trace a vertical line down from those intersection points to see the x-value.

x-value represents time in seconds.
y-value represents height in feet.