bowenescamilla
Answered

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A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground? How high will the ball go?

Sagot :

darth666

[tex]f(t)=h=-16t^2+20t+6 [/tex]

a=-16, b=20, c =6

Δ=[tex]b^2-4ac=20^2-4*(-16)*6=400+384=784[/tex]

[tex]\sqrt{Δ}=\sqrt{784}=28[/tex]

max:

[tex]p=-\frac{b}{2a}=-\frac{20}{-32}=\frac{20}{32}=\frac{5}{8}[/tex]

max heigh: [tex]h=-16(\frac{5}{8})^2+20*\frac{5}{8}+6=-16\frac{25}{64}+\frac{100}{8}+6[/tex]

[tex]h=-\frac{25}{4}+\frac{50}{4}+6=\frac{-25+50+24}{4}=\frac{49}{4}[/tex]

[tex]h=\frac{49}{4}[/tex] at [tex]t=\frac{5}{8}s[/tex]


will fall down at t:

[tex]t=\frac{-20-28}{-32}=\frac{-48}{-32}=\frac{3}{2}=1.5[/tex]

[tex]t=1.5s[/tex]

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