SOLUTION: 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

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Question 141576This question is from textbook nelson mathematics 10
: 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.
a) How long does it take for the rocket to pass a window 10 m above the ground?
b) When does the rocket hit the ground?
c)What is the max height the rocket reaches above the roof of the school? This question is from textbook nelson mathematics 10

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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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.
:
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+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation a=-5; b=23; c=5
t+=+%28-23+%2B-+sqrt%2823%5E2+-+4+%2A+-5+%2A+5+%29%29%2F%282%2A-5%29+
t+=+%28-23+%2B-+sqrt%28529+-+%28-100%29+%29%29%2F%28-10%29+
t+=+%28-23+%2B-+sqrt%28529+%2B+100+%29%29%2F%28-10%29+
t+=+%28-23+%2B-+sqrt%28629%29%29%2F%28-10%29+
Two solutions, but we only want the positive one:
t+=+%28-23+-+25.08%29%2F%28-10%29
t = %28-48.08%29%2F%28-10%29
t = + 4.8 sec the rocket will pass a window 10' from the ground
:
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 = %28-23%29%2F%282%2A-5%29
t = %28-23%29%2F%28-10%29
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%28+300%2C+200%2C+-4%2C+6%2C+-10%2C+50%2C+-5x%5E2%2B23x%2B15%29+
;
did all this make sense to you? any questions?