Question

@kirbykirby Here you go :)

Answer 1

For a) you are basically solving for time, that is you are solving for your \(t\) variable. They give you \(s=82\) and \(v=60\) from the problem. So
\(H(t)=-16t^2+60t +82\)
\(-16t^2+60t+82-H(t)=0\)

Now, you solve for t using the quadratic formula. Realize though that \(82-H(t)\) is your "c" term in the quadratic formula.
\[ t=\frac{-60\pm\sqrt{60^2-4(-16)(82-H(t))}}{2(-16)}\\
\, \\
t=\frac{-60\pm\sqrt{3600+5248-64H(t)}}{-32}\]
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b) You know that the square root can only give you a positive quantity. Furthermore, the vertex of a quadratic function (parabola) occurs when the discriminant \(b^2-4ac\)=0.

So,
\(3600+5248-64H(t)=0\)

\(-64H(t)=-3600-5248\)

\(H(t)=138.25\). The maximum height is 138.25 feet
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c) Solve \( H(t) = g(t)\)

\(-16t^2+60t+82 = 10+63.8t\\
\, \\
-16t^2-3.8t+72=0\)

Use the quadratic equation again:
\[ t=\frac{-(-3.8)\pm\sqrt{(-3.8)^2-4(-16)(72)}}{2(-16)}\\

t=-2.2434\text{ or } 2.0059 \]
Only take the positive solution as it makes physical sense., 2.0059 seconds. This solution means that after 2.0059 seconds, the 2 objects will collide! Because recall that the solution of these two functions is the point where the graphs intersect, and the graphs here represent the path taken by these objects, so the solution is their intersection point. So if the graphs intersect, the objects intersect (and thus collide).
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d) This is referring back to a solution in b). You know that the maximum of function occurs at H(t) = 138.25. So, what time does this occurs at? Just plug in H(t) in the formula you found in a) to solve for t:
\[t=\frac{-60\pm\sqrt{3600+5248-64(138.25)}}{-32}\]
If you don't have to repress this on your calculator, realize that we found that the maximum value occurs when the discriminant was 0, so the whole square root term disappears. So,
\[t=\frac{-60}{-32}=1.875\}.

Now, the graph of a projectile motion generally looks like this:
and I indicated on the graph where the maximum occurs.
Now, since the max occurs at time 1.875 seconds, but the collision occurred at 2.0059 seconds, the collision occurs AFTER the projectile has reached the maximum. So, it collided while the projectile was going down.

Answer 2

Sorry Latex typo:
If you don't have to repress this on your calculator, realize that we found that the maximum value occurs when the discriminant was 0, so the whole square root term disappears. So,
\[t=\frac{-60}{-32}=1.875\].

Answer 3

Thank you, again!!! :) :)

Answer 4

:)