Question

Someone please explain (:

Answer 1

a.) 1.38 seconds

b.) 17.59ft

Explanation:

h(t) = -16t^2 + 22.08t + 6

if we were to graph this, the vertex of the function would be the point, which if we substituted into the function would give us the maximum height.

to find the vertex, since we are dealing with something called "quadratic form" ax^2+bx+c, we can use a formula to find the vertex

-b/2a

b=22.08

a=-16

-22.08/-16, we get 1.38 when the minuses cancel out. since our x is time, it will be 1.38 seconds

now since the vertex is 1.38, we can substitute 1.38 into the function to find the maximum height.

h(1.38)= -16(1.38)^2 + 22.08t + 6 -----> is maximum height.

approximately = 17.59ft -------> calculator used, and rounded to 2 significant figures.

for c the time can be equal to (69+sqrt(8511))/100, as the negative version would be incompatible since we are talking about time. or if you wanted a rounded decimal, approx 1.62 seconds.

Answer 2

a)
`h(t)=-16(t-0.69)^2)+13.62`

b)The maximum height is 13.62 feet.

c) 1.84 seconds

Explanation:

The function that models the height in feet of the hammer above the ground is

`h(t)=-16t^2+22.08t+6`

Factor -16 as shown;

`h(t)=-16(t^2-1.38t)+6`

Add and subtract the square of half the coefficient of t.

`h(t)=-16(t^2-1.38t+(-0.69)^2)--16(-0.69)^2+6`

`h(t)=-16(t^2-1.38t+(-0.69)^2)--16(-0.69)^2+6`

Rewrite the first three terms as a perfect square trinomial;

`h(t)=-16(t-0.69)^2)+13.62`

This function is now in the form;

`h(t)=a(t-h)^2)+k`

where (h,k)=(0.69,13.62)

This implies that; the hammer reached its maximum height at t=0.69 seconds.

b) The maximum height is the y-value of the vertex.

The maximum height is 13.62 feet.

c) To answer this question we need to find the difference between the x-intercepts.

`h(t)=-16t^2+22.08t+6`

a=-16,b=22.08,c=6

`t_2-t_1=√((t_1+t_2)^2-4t_1t_2)`

`t_2-t_1=\sqrt{((-b)/(a))^2-4((c)/(a))}`

`t_2-t_1=\sqrt{((-22)/(-16))^2-4((-6)/(-16))  }=1.84`

Hence the hammer was in the air for 1.84 seconds