Respuesta :
General Idea:
When t represents the time, h(t) represent the height of ball in meters after t seconds and you have an function of the form [tex] h(t) = at^2 + bt + c [/tex], and if [tex] a < 0 [/tex], then this function will reach maximum height at [tex] t=\frac{-b}{2a} [/tex].
Since time cannot be positive, [tex] t\geq 0 [/tex].
When ball is above the ground, [tex] h(t)>0 [/tex].
When the ball hit the ground [tex] h(t) =0 [/tex]
Applying the concept:
Setting up the function equal to zero to find the time it takes for the ball hit the ground.
[tex] -5t^2+9t+80 =0\\ \\ Here\; a=-5; \; b=9; \; c = 80\\ \\ Using\; Quadratic \; formula \;\\ \\ t=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ t=\frac{-9\pm \sqrt{81-4(-5)(80)}}{2(-5)} \\ \\ t=\frac{-9\pm \sqrt{81+1600}}{-10}\\ \\ t=\frac{-9+41}{-10} (or)t=\frac{-9-41}{-10}\\ \\ Since \; t \; cannot \; be \; negative\\ \\ t=\frac{-50}{-10}=5 \; seconds [/tex]
Conclusion:
The Ball will be above the ground during the time interval[tex] 0\leq t\leq 5 [/tex].
Answer:
The function h(t) = -5(t + 3.2)(t - 5) reveals that the ball's height above the ground will be 0 meters after 5 seconds.
