SOLUTION: A ball is tossed in the air from an initial height of 6 feet, it reaches its maximum height of 10 feet after 3 seconds then comes back down.then it bounces back up to a second maxi

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Question 844782: A ball is tossed in the air from an initial height of 6 feet, it reaches its maximum height of 10 feet after 3 seconds then comes back down.then it bounces back up to a second maximum height of 1 foot in 1 second
a) Assume that the motion is parabolic in each case and determine a piecewise function h(t). express all non- exact answers rounded correctly to three decimal places.
b) now assume that the whole time it is doing this, it is moving in the horizontal(call it the x direction) at a constant speed of 1.5 feet per second. Determine the particles path through space h(x(t)) as a piecewise function as it goes through the initial toss and two bounces leave all non-exact answers rounded to three decimal places.
guys I need help in home work :) i would appreciate the help
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
This problem reads as a math problem, so I will forget that I know physics.
a) Assuming that the motion is parabolic, we expect the pieces of the graph of h%28t%29 as a function of t to look like this:
The first piece has axis of symmetry t=3 seconds,
apex/vertex at h=10 feet,
and h%280%29=6 feet.
From f%28t%29=t%5E2, a flip, a vertical stretch, and a translation bring us to
h%28t%29=10%2BK%28t-3%29%5E2 as a candidate function.
That function graphs as a parabola with a vertex at t=3 seconds with h=10 feet.
We know that we need K%3C0 for the vertex to be a maximum, and that we need h%280%29=6 feet.
h%280%29=10%2BK%280-3%29%5E2=6
10%2BK%2A3%5E2=6
10%2B9K=6
9K=6-10
9K=-4
K=-4%2F9
So the first piece of the function, from t=0 to the first bounce point, where h%28t%29=0 is
h%28t%29=10-%284%2F9%29%28t-3%29%5E2
h%28t%29=0-->10-%284%2F9%29%28t-3%29%5E2=0-->10=%284%2F9%29%28t-3%29%5E2-->10%2A%289%2F4%29=%28t-3%29%5E2-->t-3=sqrt%2810%2A%289%2F4%29%29-->t-3=%283%2F2%29%2Asqrt%2810%29-->t=3%2B%283%2F2%29%2Asqrt%2810%29seconds = approx.7.743seconds .
So the first piece is
highlight%28h%28t%29=10-%284%2F9%29%28t-3%29%5E2%29} for highlight%280%3C=t%3C7.743%29seconds .

The second bounce takes the ball up to h=1 foot 1 second later,at
t=4%2B%283%2F2%29%2Asqrt%2810%29=about8.743 seconds, and parabola symmetry requires that the ball get back to h=0 in another second at t=5%2B%283%2F2%29%2Asqrt%2810%29 seconds.
The function y=1-t%5E2 goes from y%28-1%29=0 to a maximum of y%280%29=1 in 1 second.
We need the same shape, but shifted right so the vertex is at t=4%2B%283%2F2%29%2Asqrt%2810%29=8.743 instead of t=0 .
So highlight%28h%28t%29=1-%28t-8.743%29%5E2%29 for highlight%287.743%3C=t%3C=9.743%29 would cover the second piece of the piecewise function.

b) x%28t%29=1.5t <--> x%28t%29=%283%2F2%29%2At <--> t=%282%2F3%29x
We can express the trajectory function h%28x%29 by substituting t=%282%2F3%29x into h%28t%29 .
The t%3C=0%3C7.743 part of the piecewiese function corresponds to 0%3C=x%3C7.743%2A1.5<-->highlight%280%3C=x%3C11.615%29

and we get h=highlight%2810-%284%2F81%29%282x-9%29%5E2%29 .
For the second piece, t%3C=0%3C7.743 corresponds to
11.615%3C=x%3C=9.743%2A1.5<-->11.615%3C=x%3C=14.615
and we get

However, using the more accurate 4%2B%283%2F2%29%2Asqrt%2810%29=8.743416 value, we get
3%2A8.743416=26.230 when rounded to 3 decimal places.
So h=highlight%281-%282x-26.230%29%5E2%2F9%29 for highlight%2811.615%3C=x%3C=14.615%29 .

IN REVENGE
for making me calculate a bunch of ugly numbers,
I want to point out that the problem is really out of this world.
The acceleration of the ball after it was tossed in the air was
2%2A%284%2F9%29=8%2F9 feet%2Fsecond%5E2 , but on my planet the acceleration of gravity is 32 feet%2Fsecond%5E2 .
On earth's moon, it would be 5.3 feet%2Fsecond%5E2 .
That ball is being tossed on a quite small satellite or asteroid, with little gravity.
after it bounces, the acceleration suddenly increases to
2 feet%2Fsecond%5E2 .
That is even stranger, because gravity should remain constant on any planet, satellite, or asteroid.
It would have been more consistent if the bounce had made the ball reach 1 foot is 1.5 seconds. Then the acceleration of gravity before and after the bounce would have been the same.