SOLUTION: What is the velocity if H = 64t - 16t ^ 2? How can i solve for H? How can i solve for t?

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Question 879720: What is the velocity if H = 64t - 16t ^ 2?
How can i solve for H?
How can i solve for t?

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1) The velocity is highlight%2864-32t%29 .
2) You do not need to solve for H ,
because you have the formula H=64t-16t%5E2 to calculate H when you are given t .
3) To solve for t you have to solve a quadratic equation.
H=64t-16t%5E2<-->16t%5E2-64t%2BH=0 ,
and the solutions to 16t%5E2-64t%2BH=0 , if there is any,
can be calculated applying the quadratic formula as
t+=+%2864+%2B-+sqrt%2864%5E2-4%2A16%2AH+%29%29%2F%282%2A16%29+-->t+=+%2864+%2B-+sqrt%2864%5E2-64H%29+%29%2F32-->t+=+%2864+%2B-+sqrt%2864%2864-H%29%29%29%2F32-->t+=+%2864+%2B-+8sqrt%2864-H%29%29%2F32-->highlight%28t+=+%288+%2B-+sqrt%2864-H%29%29%2F4%29
That could give two answers, or one, or none.
For example, H=0 when t=0 and when t=4 ;
H=64 for t=2 only,
and H=70 does not happen for any real value of t .

The explanations for the answers above depend on why you need the answers, and what courses you have taken and/or are taking. It is different for Calculus than for Algebra 2 or Pre-calculus. It is different for Physics with little math.

For Physics, you should know that H=64t-16t%5E2 represents
the height in feet after t seconds
of an object shot up from the ground with an initial upwards velocity of 64 feet per second,
on Earth, which pulls down on objects with an acceleration of 32 ft%22%2F%22s%5E2 .

For Calculus, I would tell you that the velocity is the derivative of H%28t%29 , dH%2Fdt .

I would give more explanations if I knew which ones you need and can use.