SOLUTION: I would really appreciate it if I could get some help on this question, thanks.
The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the init
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Quadratic Equations and Parabolas
-> SOLUTION: I would really appreciate it if I could get some help on this question, thanks.
The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the init
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Question 137457: I would really appreciate it if I could get some help on this question, thanks.
The path of a falling object is given by the function s=-16t^2+vo+s0 where vo represents the initial velocity in ft/sec and so represents the initial height in feet. Also, s represents the height in feet of the object at any time,t,which is measured in seconds.
If a rock is thrown upward with the initial velocity of 40ft per second from the top of a 30ft building.
s= -16t^2+vot+so I figured out that much, then it asks after how many seconds will the graph reach maximum height?What is the maximum height? I just need help setting up the equation to solve for height.... The question comes from my online algebra class at aiu. HELP PLEASE. thanks, alaska Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! Given: and...
Find the time, t, of the maximum height in seconds and the maximum height in feet.
Substitute the given values of and into the equation.
To find the value of t at the maximum height, you are really looking for the t-coordinate of the vertex of the parabola described by this quadratic equation.
This is given by:
The a and b come from the general form of the quadratic equation: but in your problem, the independent variable is t rather than x, where a = -16 and b = 40, so... seconds.
The maximum height is reached in 1.25 seconds.
To find the maximum height, just substitute t = 1.25 into the equation and solve for s. feet.
The maximum height reached is 55 feet.
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