SOLUTION: I do not understand the process to solve this problem:
A model rocket is launched with an initial upward velocity of 50 m/s . The rocket's height h (in meters) after t seconds i
Question 523948: I do not understand the process to solve this problem:
A model rocket is launched with an initial upward velocity of 50 m/s . The rocket's height h (in meters) after t seconds is given by the following.
h=50t-5t^2
Find all values of for which the rocket's height is 30 meters. Found 2 solutions by nerdybill, Alan3354:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! A model rocket is launched with an initial upward velocity of 50 m/s . The rocket's height h (in meters) after t seconds is given by the following.
h=50t-5t^2
Find all values of for which the rocket's height is 30 meters.
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simply set h to 30 and solve for t:
h=50t-5t^2
30=50t-5t^2
5t^2 + 30 = 50t
5t^2 - 50t + 30 = 0
t^2 - 10t + 6 = 0
Since you can't factor, you must apply the "quadratic formula" to get:
t = {0.641, 9.359} seconds.
.
That is, on the way UP at:
0.641 seconds it'll be at 30 meters
then again, on the way back DOWN at:
9.359 seconds
You can put this solution on YOUR website! A model rocket is launched with an initial upward velocity of 50 m/s . The rocket's height h (in meters) after t seconds is given by the following.
h=50t-5t^2
Find all values of for which the rocket's height is 30 meters.
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t = (10/2) - sqrt(100 - 24)/2 seconds going up
t = 5 - sqrt(19) =~ 0.6411 seconds ascending
t = 5 + sqrt(19) =~ 9.359 seconds coming back down.
