Question 141576
A model rocket is shot straight up from the roof of a school. The height at any time t is approximated by the model H = 15 + 23t -5t squared, where H is the height in metres and t is the time in seconds.
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Write the equation in the standard form:
h = -5t^2 + 23t + 15
:
you should recognize the 3 parts of this equation
-5x^2 is the force of gravity in m/sec, hence it is negative
23x is the upward velocity of the rocket in m/sec, hence it is positive
15 is the initial height or in this case the height of the school
:
a) How long does it take for the rocket to pass a window 10 m above the ground?
h = 10
-5t^2 + 23t + 15 = 10
-5t^2 + 23t + 15 - 10 = 0
-5t^2 + 23t + 5 = 0
Solve for t using the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation a=-5; b=23; c=5
{{{t = (-23 +- sqrt(23^2 - 4 * -5 * 5 ))/(2*-5) }}}
{{{t = (-23 +- sqrt(529 - (-100) ))/(-10) }}}
{{{t = (-23 +- sqrt(529 + 100 ))/(-10) }}}
{{{t = (-23 +- sqrt(629))/(-10) }}}
Two solutions, but we only want the positive one:
{{{t = (-23 - 25.08)/(-10)}}}
t = {{{(-48.08)/(-10)}}}
t = + 4.8 sec the rocket will pass a window 10' from the ground
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b) When does the rocket hit the ground?
The rocket hits the ground when h = 0, so we have:
-5t^2 + 23t + 15 = 0
Solve this equation using the quadratic formula just like we did above, except c = 15
You should get t ~ 5.18 sec
:
c)What is the max height the rocket reaches above the roof of the school?
The max height will occur at the axis of symmetry, find that using t=-b/(2a)
In this problem remember a=-5; b=23
t = {{{(-23)/(2*-5)}}}
t = {{{(-23)/(-10)}}}
t = + 2.3 sec with me max height
;
Find the max height, substitute 2.3 for t in the equation and find h
h = -5(2.3^2) + 23(2.3) + 15
h = -5(5.29) + 52.9 + 15
h = -26.45 + 52.9 + 15
h = 41.45 m is the max height
:
A graph illustrates all this 
{{{ graph( 300, 200, -4, 6, -10, 50, -5x^2+23x+15) }}}
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did all this make sense to you? any questions?