Question 688979: The equation below describes the height "h" in feet of a bullet fired from a gun up into the air "t" seconds after the bullet was fired. How high will it go, to the nearest foot? I am looking for the numerical answer to this riddle!
h=32.2t^2+644t
I can not wrap my head around this - any assistance would be greatly appreciated - thanks so much!!
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Unfortunately, you are on a different planet than Earth. On your planet the gravity is twice as strong as on Earth, and not only that, it is gravitational repulsion rather than attraction. So if you fire your bullet into the air, it will never stop until it reaches another celestial body that has gravity that works in the normal way.
On earth, the height function is \ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ h_o)
where the acceleration due to gravity,  , is  feet per second per second. Using your values of 644 feet per second for intial velocity and zero for initial height (damn short gun!), your function would look like:
\ =\ -16.1t^2\ +\ 644t)
You should recognize this as a quadratic function where the graph is a parabola opening downward. The vertex is therefore a maximum. Use the formula for the abscissa of the vertex:
Once you have performed that calculation and determined the time at which the height is maximum, evaluate your function at that value:
\ =\ -16.1\left(t_{max}\right)^2\ +\ 644\left(t_{max}\right))
Once you have finished your calculations and have returned to Earth, write back and I'll check your work for you.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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