Question 181925
If a ball is thrown into the air with an initial velicity of 52 m/s from a height of 6 m above the ground, its height after t seconds is given by the equation s(t)= -4.9t^2 +52t +6
a) what is the time it takes to het to maximum height?
that would be the "axis of symmetry" found at
t = -b/2a
t = -52/2(-4.9)
t = -52/(-9.8)
t = 5.306 seconds
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b) what is the maximum height?
Plug the answer above into:
s(t)= -4.9t^2 +52t +6
s(5.306)= -4.9(5.306)^2 +52(5.306) +6
s(5.306)= -4.9(28.153636) + (275.912) +6
s(5.306)= -137.9528164 + 281.912
s(5.306)= 143.959 feet
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c) how long does it take the ball to get to ground level?
set s(t) to zero and solve:
s(t)= -4.9t^2 +52t +6
0= -4.9t^2 +52t +6
Applying the quadratic equation yields:
t = {-0.114, 10.726}
Toss out the negative solution.
t = 10.726 seconds
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Below is the details of the quadratic solution:
*[invoke quadratic "x", -4.9, 52, 6 ]