SOLUTION: Please help. If a rock is thrown upward with an initial velocity of 40 feet per second from the top of a 30 foot building how high is the rock after 2 seconds,how many seconds will

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: Please help. If a rock is thrown upward with an initial velocity of 40 feet per second from the top of a 30 foot building how high is the rock after 2 seconds,how many seconds will the graph reach maximum height and what is the maximum height? Can someone please help This question is from textbook

Answer by Earlsdon(6294)   (Show Source): You can put this solution on YOUR website!
To start solving this type of problem, you should know that the height (h) of an object propelled upward from a height of from the earth's surface with an initial velocity of ft./sec. is given by the function:

In this problem, ft./sec. and ft. and you want to find the height, h, when t = 2 secs., so, substitute t = 2 and...



1) The rock is feet high after 2 seconds.
The maximum height can easily be found when you realize that this quadratic function, when graphed, represents a parabola that opens downwards (negative coefficient), so you need to find the location of the vertex (maximum) of the parabola.
The independent variable is t (time) so the t-coordinate of the vertex is given by:
where the a and b come from the standard form of the quadratic equation:, so, in this problem, a = -16 and b = 40


secs.
The maximum height occurs at time seconds. Now substitute this value of t into the quadratic function to find the height of the rock at time t = 1.25.

feet.
The maximum height attained by the rock is feet.

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