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

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) (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 as a function of to look like this:
The first piece has axis of symmetry seconds,
apex/vertex at feet,
and feet.
From , a flip, a vertical stretch, and a translation bring us to
as a candidate function.
That function graphs as a parabola with a vertex at seconds with feet.
We know that we need for the vertex to be a maximum, and that we need feet.

So the first piece of the function, from to the first bounce point, where is

-->-->-->-->-->-->seconds = approx.seconds .
So the first piece is
} for seconds .

The second bounce takes the ball up to foot 1 second later,at
seconds, and parabola symmetry requires that the ball get back to in another second at seconds.
The function goes from to a maximum of in 1 second.
We need the same shape, but shifted right so the vertex is at instead of .
So for would cover the second piece of the piecewise function.

b) <--> <-->
We can express the trajectory function by substituting into .
The part of the piecewiese function corresponds to <-->

and we get .
For the second piece, corresponds to
<-->
and we get

However, using the more accurate value, we get
when rounded to 3 decimal places.
So for .

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

, but on my planet the acceleration of gravity is .
On earth's moon, it would be .
That ball is being tossed on a quite small satellite or asteroid, with little gravity.
after it bounces, the acceleration suddenly increases to

.
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.
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