SOLUTION: a successful juggler tosses props upward with one hand and catches them by moving her other hand in an intercepting parabolic curve. the height of an orange used in jojoe's jugglin

Question 936137: a successful juggler tosses props upward with one hand and catches them by moving her other hand in an intercepting parabolic curve. the height of an orange used in jojoe's juggling act can be modeled by the function h(t) = -16t^2 + 27t + 5, where t represents time in seconds and h(t) is height in feet above the ground. jojoe's right hand moves in a parabolic curve to catch the falling orange. the path of her hand is represented by the function j(t) = 6t^2 - 10t + 3. find the time, to the nearest tenth of a second, at which jojoe catches her flying fruit.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
she catches the fruit when:
h(t) = j(t)
-16t^2 + 27t + 5 = 6t^2 - 10t + 3
27t + 5 = 22t^2 - 10t + 3
5 = 22t^2 - 37t + 3
0 = 22t^2 - 37t - 2
.
Applying the "quadratic formula" we get:
t = {1.73, -0.05}
we throw out the negative solution leaving:
t = 1.7 seconds
.
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=1545 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.73423836395607, -0.0524201821378885. Here's your graph:

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