- #1
illjazz
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Homework Statement
If a stone is thrown vertically upward from the surface of the moon with a velocity of 10m/s, its height (in meters) after t seconds is h = 10t - 0.83t^2.
a) What is the velocity of the stone after 3s?
b) What is the velocity of the stone after it has risen 25m?
Homework Equations
h = 10t - 0.83t^2
The Attempt at a Solution
a) was easy and I got that. I took the derivative of h and used that velocity function (assuming h is the position function), to get v(3) = 5.02 m/s, which is correct according to my book.
b) is giving me a load of trouble though. I've tried all kinds of equations, like the following:
h = 10t - 083t^2 = 25 and solving for t
v(t) = 10 - 1.66t = 25 and solving for t
and a few others that don't make sense, like plugging 25 into the function for t, but since this is a function of distance depending on time elapsed, the input variable expected by the function is a value of seconds. In other words, I understand I can't just plug in a distance of 25m into a function that uses an integer of seconds to give back distance travelled.
The answer in the back of the book is [tex]\sqrt{17}[/tex] which is equal to about 4.123...
Nothing I've tried so far has given me back that number.
Now, since the original position function h has as its output a distance of meters, it would make sense to solve for t where h = 25, and then taking the result of that, call it x, and solving for t with v(t) = x, no? I'm pretty sure I've tried that and it still hasn't worked out.
What am I missing?