Mark is in a deep hole looking for treasure. he is standing 8 feet below the surface. he throws an old watch he found with an initial upward velocity of 23 ft/sec. how long until it lands outside the hole, having gone up and back down? correct is b 0.8 seconds

The height of the old watch as a function of time can be modeled by the equation
     h(t) = –16t^2 + 23t - 8

We set this equation equal to zero because the height the old watch will land outside the hole is zero:
     0 = –16t^2 + 23t - 8

Since we have a quadratic equation, we can calculate for t by using the quadratic formula
     x = [-b ± sqrt(b^2 - 4ac)] / 2a 
     t = {-23 ± sqrt[23^2 - 4(-16)(-8)]} / 2(-16)
     t = [-23 ± sqrt(17)] / (-32)
     t =  0.590 seconds or  0.848 seconds

The two t values show that the old watch will land outside the hole once at 0.590 seconds and again at 0.848 seconds. Therefore, having gone up and back down, it will land at 0.848 seconds. 

Bruh i don’t know the answer tbh but you can google it