a trampoline artist bounces on a trampoline her height above the ground is modelled by the function h=-4.9(t-1)²+6.3 "h" measures the height above the ground in meters and "t" measures the time after she leaves the trampoline in seconds. corresponding question above.
what is her maximum height how long after she leaves trampoline does she reach max height how long is she in air how high off the ground is trampoline
answer found 6,3m and in t = 1 she will reach her max height is it correct
2)how long is she in the air time between she leaves the trampoline until she will reach it again
answer found 1,4
3)how high is the trampoline off the ground , i need answer for this one plz tnx.
1 Answer | Add Yours
Answers: a) 6.3m b) 1.0s c) 1.4m
Given the quadratic, `h(t) =-4.9text(m)/text(s^2)(t-1text(s))^2+6.3text(m)` , you are to determine the following:
1. Maximum height.
2. Time to maximum height.
3. Height of trampoline.
The quadratic graphs a parabola. Knowing a bit about its characteristics will help understand questions like this.
This equation is already in vertex-format (h,k):
`y(t) = a(t-h)^2+k`
Where h is time to maximum height, and k is the maximum. (This is the case because a is negative - otherwise k is the minimum).
The time to the maximum is 1.0 seconds, and the heighest point is 6.3 m
So far so good. The height of the trampoline is another story. Let's take a look at the graph and see what we find:
The graph verifies our first answers! Notice that the graph intersects the y-axis. Remember that y represents height and x represents time. The y-intercept happens when x is zero. In terms of the problem, this is where the trampoline artist is at time=0, probably at the trampoline.
So, evaluate h(0) and you will have your last answer!
`h(0) = 1.4text(m)`
The trampoline must be 1.4 m high off the ground. Now as the graph and equation suggests, at 2s, either the trampoline artist misses the trampoline or the graph starts over again - you decide :) Good job!
Answers: a) 6.3 m b) 1.0s c) 1.4m
We’ve answered 315,819 questions. We can answer yours, too.Ask a question