SOLUTION: . A baseball is hit so that its height in feet after t seconds is
s(t) = -16t2 + 44t + 4.
a) How high is the baseball after 1 seconds?
b) Find the maximum height of
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-> SOLUTION: . A baseball is hit so that its height in feet after t seconds is
s(t) = -16t2 + 44t + 4.
a) How high is the baseball after 1 seconds?
b) Find the maximum height of
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Question 669765: . A baseball is hit so that its height in feet after t seconds is
s(t) = -16t2 + 44t + 4.
a) How high is the baseball after 1 seconds?
You can put this solution on YOUR website! Given, the height of the ball is
(1) h(t) = -16t^2 +44t +4
At t = 1 we get
(2) h(1) = -16*1 +44 +4 or
(3) h(1) = 32 ft
Using derivative calculus, we get the time when the ball reaches the point of maximum height is given by
(4) -2*16*t + 44 = 0 or
(5) t = 11/8 sec
The maximum height is
(6) h(11/8) = -16*(11/8)^2 +44*(11/8) +4 or
(7) h(11/8) = -16*121/64 +121/2 +4 or
(8) h(11/8) = 121*(1/2 - 1/4) +4 or
(9) h(11/8) = 121/4 + 4 or
(10) h(11/8) = (121 + 16)/4 or
(11) h(11/8) = 137/4 or
(12) h(11/8) = 34 1/4 ft
Answers: After 1 sec the baseball is at 32' height and reaches a maximum height of 34'-3" at 1 and 3/8 sec.
You can put this solution on YOUR website! . A baseball is hit so that its height in feet after t seconds is
s(t) = -16t2 + 44t + 4.
a) How high is the baseball after 1 seconds?
b) Find the maximum height of the baseball.
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without derivatives:
The vertex of the parabola is at t = -b/2a = -44/-32
t = 11/8 seconds.
The calculations are the same.
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I would also use the derivative, but I don't know if you are familiar with them.
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