Answer:
Alright, we can express the baseball's motion with an equation like:
h(x) = a*x^2 + b*x + c
Here, x denotes time, while h(x) indicates height.
Let’s construct this:
The acceleration is:
a(t) = a
For velocity, integrating over time results in:
v(x) = a*x + v0
Where v0 signifies the initial vertical velocity.
Subsequently, we can determine position or height by integrating once more:
h(x) = a*x^2 + v0*x + h0
Here, h0 is the initial height.
<pthus our="" equation="" is:="">
h(x) = a*x^2 + v0*x + h0.
<pexamining the="" table:="">
When x = 0s, h(0s) = 6ft
<pthus:>
h(0s) = a*0s^2 + v0*0s + h0 = 6ft
h0 = 6ft.
It’s also noted that:
h(2s) = h(4s)
<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.</p"><pin a="" standard="" quadratic="" function:="">
a*x^2 + b*x + c
The symmetry line is given by:
x = -b/2a
<pin this="" instance:="">
b = v0
a = a
<ptherefore we="" derive:="">
3s = -v0/(2*a)
v0 = -3s*(2a)
<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">
h(x) = a*x^2 - 6s*a*x + 6ft
<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>
h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft
<pthus our="" resulting="" equation="" reads:="">
h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft
b) The height after 5 seconds is expressed as:
h(5s) = (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft
</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>
