Respuesta :
Answer:
Since time can't be negative, the only possible solution is 0.72 s.
Step-by-step explanation:
When the ball hits the ground its height is equal to 0, so we must find a value for "t" that results in h = 0. We have:
[tex]-16*t^2 + 6*t + 4 = 0\\t_{1,2} = \frac{-6 \pm \sqrt{(6)^2 - 4*(-16)*(4)}}{2*(-16)}\\t_{1,2} = \frac{-6 \pm \sqrt{292}}{-32}\\t_{1,2} = \frac{-6 \pm 17.1}{-32}\\t_1 = \frac{-6 -17.1}{-32} = 0.72\\t_2 = \frac{-6 + 17.1}{-32} = -0.35\\[/tex]
Since time can't be negative, the only possible solution is 0.72 s.
Answer:
.72
Step-by-step explanation:
I got it right on the test!!!
