`h=-4.9(t-1)^2+6.3`

The given function is in vertex form of the parabola `y=a(x-k)^2+k` , where (h,k) is the vertex.

This indicates that the vertex of the function is (1, 6.3).

Since the sign of *`a` * in the function is negative, the vertex is the maximum point of the parabola.

**Hence, the maximum height of the trampoline artist is 6.3m above the ground and it takes her 1 second to reach the maximum height.**

To determine the amount of time that she is in the air, set the height above the ground equal to zero (h=0).

`h=-4.9(t-1)^2+6.3`

`0=-4.9(t-1)^2+6.3`

`-6.3=-4.9(t-1)^2`

`(-6.3)/(-4.9)=(t-1)^2`

`1.29=t^2-2t+1`

Set one side equal to zero.

`0=t^2-2t+1-1.29`

`0=t^2-2t-0.29`

Apply quadratic formula to solve for t.

`t=(-b+-sqrt(b^2-4ac))/(2a)=(-(-2)+-sqrt((-2)^2-4*1(-0.29)))/(2*1) = (2+-sqrt5.16)/2=(2+-2.27)/2`

`t = (2+2.27)/2=2.14` and `t=(2.-2.27)/2=-0.14`

Take only the positive value of t. This the amount of time the it takes the artist to reach the ground.

**Hence, the artists is the air for 2.14 seconds.**

Note that t measures the time after the artist leave the trampoline. So to determine the height of the trampoline above the ground, set t=0.

`h=-4.9(t-1)^2+6.3`

`h=-4.9(0-1)^2+6.3=-4.9*1+6.3=1.4`

**Thus, the trampoline is 1.4 meters above the ground.**

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