Question 844782
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(t)}}} as a function of {{{t}}} to look like this:
{{{drawing(300,300,-0.1,0.9,-0.1,0.9,
graph(300,300,-0.1,0.9,-0.1,0.9,0.8-5(x-0.4)^2),
locate(0.03,0.87,h(t)),line(-0.07,0.84,-0.01,0.84),
locate(0.87,0.07,t),line(0.85,-0.04,0.9,-0.04)
)}}} The first piece has axis of symmetry {{{t=3}}} seconds,
apex/vertex at {{{h=10}}} feet,
and {{{h(0)=6}}} feet.
From {{{f(t)=t^2}}}, a flip, a vertical stretch, and a translation bring us to
{{{h(t)=10+K(t-3)^2}}} 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<0}}} for the vertex to be a maximum, and that we need {{{h(0)=6}}} feet.
{{{h(0)=10+K(0-3)^2=6}}}
{{{10+K*3^2=6}}}
{{{10+9K=6}}}
{{{9K=6-10}}}
{{{9K=-4}}}
{{{K=-4/9}}}
So the first piece of the function, from {{{t=0}}} to the first bounce point, where {{{h(t)=0}}} is
{{{h(t)=10-(4/9)(t-3)^2}}}
{{{h(t)=0}}}-->{{{10-(4/9)(t-3)^2=0}}}-->{{{10=(4/9)(t-3)^2}}}-->{{{10*(9/4)=(t-3)^2}}}-->{{{t-3=sqrt(10*(9/4))}}}-->{{{t-3=(3/2)*sqrt(10)}}}-->{{{t=3+(3/2)*sqrt(10)}}}seconds = approx.{{{7.743}}}seconds .
So the first piece is
{{{highlight(h(t)=10-(4/9)(t-3)^2)}}}} for {{{highlight(0<=t<7.743)}}}seconds .
 
The second bounce takes the ball up to {{{h=1}}} foot 1 second later,at
{{{t=4+(3/2)*sqrt(10)=about8.743}}} seconds, and parabola symmetry requires that the ball get back to {{{h=0}}} in another second at {{{t=5+(3/2)*sqrt(10)}}} seconds.
The function {{{y=1-t^2}}} goes from {{{y(-1)=0}}} to a maximum of {{{y(0)=1}}} in 1 second.
We need the same shape, but shifted right so the vertex is at {{{t=4+(3/2)*sqrt(10)=8.743}}} instead of {{{t=0}}} .
So {{{highlight(h(t)=1-(t-8.743)^2)}}} for {{{highlight(7.743<=t<=9.743)}}} would cover the second piece of the piecewise function.
 
b) {{{x(t)=1.5t}}} <--> {{{x(t)=(3/2)*t}}} <--> {{{t=(2/3)x}}}
We can express the trajectory function {{{h(x)}}} by substituting {{{t=(2/3)x}}} into {{{h(t)}}} .
The {{{t<=0<7.743}}} part of the piecewiese function corresponds to {{{0<=x<7.743*1.5}}}<-->{{{highlight(0<=x<11.615)}}}
{{{h(x(t))=10-(4/9)((2/3)x-3)^2}=10-(4/9)((2x-9)/3)^2=10-(4/81)(2x-9)^2}}}
and we get {{{h=highlight(10-(4/81)(2x-9)^2)}}} .
For the second piece, {{{t<=0<7.743}}} corresponds to
{{{11.615<=x<=9.743*1.5}}}<-->{{{11.615<=x<=14.615}}}
and we get
{{{h(x(t))=1-(t-8.743)^2=1-((2/3)x-8.743)^2=1-(2x-3*8.743/3)^2=1-(2x-26.229)^2/9}}}
However, using the more accurate {{{4+(3/2)*sqrt(10)=8.743416}}} value, we get
{{{3*8.743416=26.230}}} when rounded to 3 decimal places.
So {{{h=highlight(1-(2x-26.230)^2/9)}}} for {{{highlight(11.615<=x<=14.615)}}} .
 
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*(4/9)=8/9}}} {{{feet/second^2}}} , but on my planet the acceleration of gravity is {{{32}}} {{{feet/second^2}}} .
On earth's moon, it would be {{{5.3}}} {{{feet/second^2}}} .
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/second^2}}} .
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.