The height h = -8t² + 32t +6
The ball hits the ground when height is zero, then:
-8t² + 32t +6 = 0
To solve this we will use the quadratic formula:
[tex]t\text{ = }\frac{-(-8)\pm\sqrt[]{32^2-4(-8)(6)}}{2(-8)}=\frac{8\pm\sqrt[]{1024+192}}{-16}=\frac{8\pm\sqrt[]{1216}}{-16}=\frac{8\pm34.8712}{-16}[/tex]t1 = (8 + 34.8712)/(-16) = -2.68
t2 = (8 - 34.8712)/(-16) = 1.67
Since the time cannot be negative, we choose 1.67 as the solution
Answer:
1.67 seconds
