Question 179599: A rock is thrown from the top of a building. The distance is feet, between the rock and the ground "t" seconds after it is thrown is given by d=16t^2+64X+128
what is the maximum height that the rock reaches?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Your problem statement seems to have some anomalies in it!
The standard formula for the height (as a function of time) of an object propelled upwards from an initial height of  with an initial velocity of  is given by:
 where g (the constant of acceleration due to gravity) is 32 ft/sec^2, so your formula should look like:
Your formula also shows two different varaibles. t and x.
So, back to the problem but starting with the correct formul, (the d in your problem will work just as well as the h(t)) we have:
 First, we'll find the time, t, at which the height, d, is at its maximum. The graph of the quadradtic equation that we have here is a parabola that opens downward (signified by the negative first term) so its maximum will occur at the vertex of the parabola, and this is given by:
 where a = -16 and b = 64, so...
 }
 secs.
To find the maximum height, we substitute the time, t=2secs into the original equation and solve for d.
 feet.
The rock will reach a maximum height of 192 feet.
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