SOLUTION: h= -16t2 +80t + 50 Use this position polynomial to calculate the following: 1.The height of the object after 2 seconds 2.The height of the object after 5 seconds 3.The max

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Question 316805: h= -16t2 +80t + 50
Use this position polynomial to calculate the following:
1.The height of the object after 2 seconds
2.The height of the object after 5 seconds
3.The maximum height of the object
4.How long the the object will take to reach the ground?
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
h= -16t2 +80t + 50
Use this position polynomial to calculate the following:
1.The height of the object after 2 seconds
h= -16(2)^2 +80(2) + 50
h= -16(4) +160 + 50
h= -64 +160 + 50
h = 146
2.The height of the object after 5 seconds
h= -16(5)^2 +80(5) + 50
h= -16(25) +400 + 50
h= 50
3.The maximum height of the object
axis of symmetry:
t = -b/(2a) = -80/(-32) = 2.5
h= -16(2.5)^2 +80(2.5) + 50
h = 150
4.How long the the object will take to reach the ground?
set h to zero solve for t
0= -16t2 +80t + 50
Apply the quadratic formula to get:
t = {-0.562, 5.562}
Toss out the negative solution leaving:
t = 5.562
Details of quadratic to follow:

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=9600 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.561862178478973, 5.56186217847897. Here's your graph: