The height of a baseball, t seconds after it is hit, is given by h(t) = -16t^2 + 80t + 5. Find the maximum height the ball reaches. Please show me how you solved it.
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h(t) = -16t^2 + 80t + 5
h'(t) = -32t + 80 = 0
t = 2.5 s
h(2.5) = 180 meters
h'(t) = -32t + 80 = 0
t = 2.5 s
h(2.5) = 180 meters
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You have to plug in different numbers (0, 1, 2, 3, etc.) into the variable "t". Make a chart, where x is the plug-in number and y is the result of plugging it into the equation. It is a parabolic equation, because the baseball goes up and comes down in a curve.
For example:
x | y
0 | 5
1 | 69
2 | 101
3 | 101
4 | 69
and so on, and so on. At some point, the height will decrease since the ball has to fall back to Earth. So the maximum height the ball reaches is 101 feet or meters or whatever the unit is.
I hope this helps!
For example:
x | y
0 | 5
1 | 69
2 | 101
3 | 101
4 | 69
and so on, and so on. At some point, the height will decrease since the ball has to fall back to Earth. So the maximum height the ball reaches is 101 feet or meters or whatever the unit is.
I hope this helps!
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^Ahh beat me to it. The answer is above me, you have to take the derivative of that function. By setting it equal to zero, that point shows you where the slope is zero therefore is either a max or min, in this case it is a maximum. You will get time, so plug it back into original equation to get your height.
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find the values of t where h(t)=0, find the average of the two values, and that will be the time it takes for the ball to reach its maximum height. Substitute that value in for t to find the maximum height the ball reaches.
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You have to integrate the equation
so youll get ( im thinking its suppose to be positive)
= 32t+80
solve for t
then once you find an answer for t
plug that answer back in he original equation
so youll get ( im thinking its suppose to be positive)
= 32t+80
solve for t
then once you find an answer for t
plug that answer back in he original equation