SOLUTION: the function -16t^2 + 64t + 192 give the height S, in ft, of a model water rocket launched with a velocity of 64 ft/second from a hill that is 192 ft high. a) determine how long it
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Question 108154This question is from textbook college algebra graphs & models
: the function -16t^2 + 64t + 192 give the height S, in ft, of a model water rocket launched with a velocity of 64 ft/second from a hill that is 192 ft high. a) determine how long it will take the rocket to reach the ground, b) find the interval on which the height of the rocket is greater than 240 ft.
This question is from textbook college algebra graphs & models
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
the function -16t^2 + 64t + 192 give the height S, in ft, of a model water rocket launched with a velocity of 64 ft/second from a hill that is 192 ft high.
:
a) determine how long it will take the rocket to reach the ground,
The height of the ground is 0 (naturally) so we have:
-16t^2 + 64t + 192 = 0
:
Simplify and change the signs, divide by -16:
(it's easier to factor, when the coefficient of t^2 is positive)
+t^2 - 4t - 12 = 0
Factors toP
(t - 6)(t + 2) = 0
:
t = + 6 seconds to reach the ground (neg solution is not used)
:
:
b) find the interval on which the height of the rocket is greater than 240 ft.
Find the times when the rocket is at 240 ft
:
-16t^2 + 64t + 192 = 240
:
-16t^2 + 64t + 192 - 240 = 0
-16t^2 + 64t - 48 = 0
:
Simplify and change the signs, divide by -16:
t^2 - 4t + 3 = 0
Factors to:
(t - 3)(t - 1) = 0
Two solutions:
t = +1 sec rocket reaches 240 ft on the way up
and
t = +3 sec rocket reaches 240 ft on the way down
:
The interval between 1 and 3 sec the rocket was above 240 ft
:
:
Did this makes sense to you? Not that hard, right?
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