On top of a hill, a rocket is launched from a distance 80 feet above a lake. The rocket will fall into the lake after its engine burns out. The rocket's height, h, in feet above the surface of the lake, is given by the equation, h = -16t 2 + 64t + 80, where t is time in seconds. The maximum height of the rocket is _______ feet. What is the max height and how do you figure out the problem.
+37170 You can graph the quadratic to see the rocket's path....you will see it peaks out at 144 feet
Or, if you know how, you can take the DERIVATIVE of the quadratic (which results in an equation giving you the SLOPE of the original quadratic) and set = 0 (this would be when the rocket's path is at it's apex) to find 't' ...then substitute this 't' into the original equation
Derivative = -32t + 64 = 0 results in 't' = 2 seconds
Substitute 't' = 2 into the original equation:
h = -16(2)^2 +64(2) + 80 = 144 feet max (pretty wimpy rocket !)
Or, you can 'trial and error' it to find the 't' that gives you greatest 'h'
+37170 You can graph the quadratic to see the rocket's path....you will see it peaks out at 144 feet
Or, if you know how, you can take the DERIVATIVE of the quadratic (which results in an equation giving you the SLOPE of the original quadratic) and set = 0 (this would be when the rocket's path is at it's apex) to find 't' ...then substitute this 't' into the original equation
Derivative = -32t + 64 = 0 results in 't' = 2 seconds
Substitute 't' = 2 into the original equation:
h = -16(2)^2 +64(2) + 80 = 144 feet max (pretty wimpy rocket !)
Or, you can 'trial and error' it to find the 't' that gives you greatest 'h'
+37170 Oh yah, another way
hmax = c - (b^2)/(4a)
hmax = 80 - (64^2)/(-64)
= 80 + 64 = 144 feet (STILL a wimpy rocket !)
