SOLUTION: please help me solve this problem: A rocket is fired straight up from the ground at a time t= 0 with an initial speed of 560 feet per second. The rocket's distance of h feet above
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Question 319109: please help me solve this problem: A rocket is fired straight up from the ground at a time t= 0 with an initial speed of 560 feet per second. The rocket's distance of h feet above the ground t seconds later h=560t-16t^2 At what time will the rocket be 3, 136 feet above the ground?
Found 2 solutions by nyc_function, solver91311:
Answer by nyc_function(2741) (Show Source): You can put this solution on YOUR website!
In other words, let h = 3136 and solve for t.
3136 = 560t - 16t^2
16t^2 - 560t + 3136 = 0
You have a quadratic equation here.
Can you solve this equation?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The more general formula is:
\ =\ -16t^2\ +\ v_ot\ +\ h_o)
Where 
is the height as a function of time, 
, 
is the initial velocity, and 
is the initial height.
For your problem the rocket was launched from the ground so 
, and 
fps.
We need to know the value of that makes \ =\ 3,136)
feet.
Divide through by 
This quadratic factors quite neatly, yielding two positive roots. I'll leave it as an exercise for the student to figure out the significance of two answers, both apparently valid.
John
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