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

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:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=2821.6 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.114156624968866, 10.726401522928. Here's your graph:

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