A ball is thrown into the air with an upward velocity of 48ft/s. It's height h in feet after t seconds is given by the function h=-16t^+48t+8. In how many seconds will the ball hit the ground?

I got 1.5 seconds put when I plug it back into the equation, I get it wrong. Thanks for helping!!

To find the time it takes for the ball to hit the ground, we need to find the value of t when the height h is equal to zero.

Given that the height function is h = -16t^2 + 48t + 8, we can set h equal to zero and solve for t:

0 = -16t^2 + 48t + 8

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = -16, b = 48, and c = 8.

Plugging these values into the quadratic formula, we have:

t = (-48 ± √(48^2 - 4(-16)(8))) / (2(-16))

Simplifying further:

t = (-48 ± √(2304 + 512)) / (-32)

t = (-48 ± √(2816)) / (-32)

t = (-48 ± 53.066) / (-32)

Now, we have two possible solutions:

t1 = (-48 + 53.066) / (-32) ≈ 0.169 seconds
t2 = (-48 - 53.066) / (-32) ≈ -3.169 seconds

Since time cannot be negative in this context, we disregard t2.

Therefore, the ball will hit the ground approximately 0.169 seconds after it was thrown.

-16t^2 + 48t + 8 = 0

first factor out a -8, and you have

2t^2 - 6t - 1 = 0

1.5 is not a root of that quadratic.