A ball is thrown into the air with an initial upward velocity of 46 ft/s. Its height (h) in feet after t seconds is given by the function ½ = -162* + 46t +6. After how many seconds will the ball hit the ground?

Answer 1

To find out when the ball hits the ground, we need to set the height h to 0 and solve for t.

0 = -16t^2 + 46t + 6

Rearranging the equation:

16t^2 - 46t - 6 = 0

Now we can solve for t using the quadratic formula:

t = (-(-46) ± √((-46)^2 - 4*16*(-6))) / 2*16
t = (46 ± √(2116 + 384)) / 32
t = (46 ± √2500) / 32
t = (46 ± 50) / 32

There are two possible solutions:

t = (46 + 50) / 32 = 96 / 32 = 3 seconds
t = (46 - 50) / 32 = -4 / 32 = -0.125 seconds

Since time cannot be negative, the ball hits the ground after 3 seconds.