Answer:
The answer is 4.06 seconds.
Step-by-step explanation:
1. We need to identify the value of h in the given equation. h stands for the final height of the ball. When the ball hits the ground, the height is 0, hence h=0, and now we replace
[tex]0=-16t^{2} +64t+4[/tex]
2. We can see this is quadratic equation, hence, we can use the quadratic equation to isolate the time " t". The quadratic formula is:
[tex]\frac{-b+-\sqrt{b^2-4ac} }{2a}\\[/tex]
where "a" is the coefficient right next to t^2, "b" is the coefficient right next to t and "c" is the coefficient that is alone. For this problem, we have a=16, b=64 and c=4
3. Now, we can replace
[tex]t=\frac{-64+-\sqrt{64^2-4(-16)(4)} }{2(-16)} \\t=\frac{-64+-65.96 }{-32} \\[/tex]
Using the plus sign (+) we get t=-0.061 sec, and using the less sign (-) we get t=4.06 seconds. We only need to take into account the positive time because there is no sense to have a negative time.
