SOLUTION: 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
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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?
b) what is the maximum height?
c) how long does it take the ball to get to ground level? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! 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
.
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
.
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
.
Below is the details of the quadratic solution:
