Simplest form of -1/2 + 4/9<br>

d) The domain would be only the positive real numbers, from 0 to 4, since we found that the ball hit the ground at t equal to 4 seconds. However, if we were talking about a quadratic function with no time involved, the domain would be all the real numbers.

Since we are working with a quadratic function which describes the ball's motion in function of the time, we need to remember the following:

- The general equation of the parabola is:

$y=ax^{2} +bx+c$

- We can calculate the coordinates of the vertex of the parabola using the following formula:

$x_{vertex}=\frac{-b}{2a}$

- Evaluating a function means replacing the variable with the given value to evaluate.

The given function is:

$h(t)=-16x^{2}+63t+4$

Where,

$a=-16\\b=63\\c=4$

Now, calculating we have:

a) What is the height of the ball after 3 seconds?

We need to evaluate the time of 3 seconds into the function, so:

$h(3)=-16(3)^{2} +63(3)+4=49feet$

So, the height of the ball after 3 seconds is 49 feet.

b) What is the maximum height of the ball?

Since the function is describing the motion of a ball thrown into the air, we can find the maximum height by finding the y-coordinate of the vertex. If the parabola opens downward or upward, the vertex will be always the highest or the lowest point of the parabola.

So, calculating the vertex we have:

$x_{vertex}=\frac{-b}{2a}=\frac{-63}{2*-16}\\\\x_{vertex}=\frac{-63}{2*-16}=\frac{-63}{-32}=1.97$

Then, replacing "x" into the equation of the parabola, we find the y-coordinate of the vertex:

$y=-16(1.97)^{2}+63(1.97)+4=-16*3.88+63*1.97+4\\y=-16*3.88+63*1.97+4=-62.08+124.11+4=66.03$

So, if the y-coordinate is 66.03, the maximum height of the ball is 66.03 feet, or 66 feet (rounded to the nearest foot).

c) When will the ball hit the ground?

We can find the time when the ball hit the ground by making equal to 0 the function and finding the roots (zeroes)

Since it's a quadratic function, we can find the zeroes using the quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

Substituting a, b and c, we have:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-63+-\sqrt{63^{2}-4*(-16)*(4)} }{2*(-16)}\\\\\frac{-63+-\sqrt{63^{2}-4*(-16)*(4)} }{2*(-16)}=\frac{-63+-\sqrt{3969+256} }{-32}\\\\\frac{-63+-\sqrt{3969+256} }{-32}=\frac{-63+-\sqrt{4225} }{-32}=\frac{-63+-(65)}{-32}\\\\t1=\frac{-63-(65)}{-32}=4\\\\t1=\frac{-63+(65)}{-32}=-0.06$

Now, since negative time does not exists, we can conclude that the ball hit the ground after 4 seconds.

d) what domain makes sense for the function?

Since the function represents the motion of the thrown ball at "t" time, the domain would be only the positive real numbers, from 0 to 4, since we found that the ball hit the ground at t equal to 4 seconds. However, if we were talking about a quadratic function with no time involved, the domain would be all the real numbers,

Note: I've attached the graph of the function.

Have a nice day!