H=-16t^2+122t+5 where h is height in feet and t is time in seconds. How long will it take the baseball to hit the ground? h=-16t^2+162t+0.2 where h is height in feet and t is time in seconds. How long will it take the golf ball to hit the ground?

Answer:

[tex]t= 7.67[/tex]

[tex]t= 10.13s[/tex]

Step-by-step explanation:

Solving (a):

[tex]H=-16t^2+122t+5[/tex]

Required: Time to hit the ground

This means that H = 0

So, we have:

[tex]0=-16t^2+122t+5[/tex]

Rewrite as:

[tex]16t^2-122t-5=0[/tex]

Solve for t using:

[tex]t= \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{122 \± \sqrt{(-122)^2 - 4*16*-5}}{2*16}[/tex]

[tex]t= \frac{122 \± \sqrt{15204}}{2*16}[/tex]

[tex]t= \frac{122 \± 123.3}{32}[/tex]

Split:

[tex]t= \frac{122 + 123.3}{32}[/tex] or [tex]t= \frac{122 - 123.3}{32}[/tex]

[tex]t= \frac{245.3}{32}[/tex] or [tex]t= \frac{-1.3}{32}[/tex]

[tex]t= 7.67[/tex] or [tex]t= -0.0406[/tex]

Time can not be negative. so: [tex]t= 7.67[/tex]

Solving (b):

[tex]h=-16t^2+162t+0.2[/tex]

Required: Time to hit the ground

This means that h = 0

[tex]0=-16t^2+162t+0.2[/tex]

Rewrite as:

[tex]16t^2-162t-0.2= 0[/tex]

Solve for t using:

[tex]t= \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{162 \± \sqrt{(-162)^2 - 4*16*-0.2}}{2*16}[/tex]

[tex]t= \frac{162 \± \sqrt{26256.8}}{32}[/tex]

[tex]t= \frac{162 \± 162.04}{32}[/tex]

[tex]t= \frac{162 + 162.04}{32}[/tex] or [tex]t= \frac{162 - 162.04}{32}[/tex]

[tex]t= \frac{324.04}{32}[/tex] or [tex]t= \frac{-0.04}{32}[/tex]

Time can not be negative. So:

[tex]t= \frac{324.04}{32}[/tex]

[tex]t= 10.13s[/tex]

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