A skydiver drops her watch as she jumps out of a plane flying at an altitude of 6,400 feet. If the equation for height as a function of time is h(t)=-16t^2+initial height where t is time in seconds and h(t) is height in feet, how many seconds will it take for the watch to hit the ground?

To hit the ground, it will take the watch:

20 seconds.

Step-by-step explanation:

Making use of the provided equation:

h (t) = - 16t ^ 2 + initial height

Which can be modified specifically for this case:

h (t) = - 16t ^ 2 + 6400 feet.

Time must be replaced a certain number of times until the value is zero, this can be done one by one, but since it would be too many iterations I will show you some examples and what you could deduce in each case.

With t = 1 second:

h (1) = - 16 (1) ^ 2 + 6400 feet = 6384 feet (only the watch has dropped 16 feet)

With t = 7 seconds:

h (7) = - 16 (7) ^ 2 + 6400 feet = 5616 feet (has fallen 784 feet)

With h = 15 seconds:

h (15) = - 16 (15) ^ 2 + 6400 ft = 2800 feet (3600 ft has fallen, it is not long)

With h = 21 seconds:

h (21) = - 16 (21) ^ 2 + 6400 feet = -656 feet (When obtaining a negative number, it is understood that the time was too long, therefore a shorter time must be taken)

With h = 20 seconds:

h (20) = - 16 (20) ^ 2 + 6400 feet = 0 feet

Since with 20 seconds the exact value of zero is obtained, this time is the exact time it would take the watch to fall to the ground, since when this time is reached the height (h) will be zero, that is, at ground level .

16 minutes is correct

The answer to your question is b’(16,14)