Question

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?

Answer 1

$f(t)=h=-16t^2+20t+6$

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

Δ=$b^2-4ac=20^2-4*(-16)*6=400+384=784$

$\sqrt{Δ}=\sqrt{784}=28$

max:

$p=-\frac{b}{2a}=-\frac{20}{-32}=\frac{20}{32}=\frac{5}{8}$

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

$h=-\frac{25}{4}+\frac{50}{4}+6=\frac{-25+50+24}{4}=\frac{49}{4}$

$h=\frac{49}{4}$ at $t=\frac{5}{8}s$

will fall down at t:

$t=\frac{-20-28}{-32}=\frac{-48}{-32}=\frac{3}{2}=1.5$

$t=1.5s$