An object is dropped from a cliff thatis 256 m above the sea. just prior to splashing into the water, how fast is it going? How long did it take to fall?

t=sqrt(2h/g)

v= gt=sqrt(2gh)

3s

14.3 m/s

To determine the speed and time it takes for the object to fall, we can use the laws of motion. Specifically, we can apply the equations of motion for a freely falling object under the influence of gravity.

The first equation we need is the equation for the height of the falling object (h) as a function of time (t):
h = (1/2) g t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time.

Rearranging the equation, we can solve for t:
t = sqrt((2h)/g)

Substituting the given height (h = 256 m) and the acceleration due to gravity (g = 9.8 m/s^2), we get:
t = sqrt((2 * 256 m) / (9.8 m/s^2))

Calculating the value, t is approximately 5.07 seconds.

Now, to determine the velocity of the object just before impact, we can use another equation of motion:
v = g * t
where v is the velocity and t is the time.

Substituting the acceleration due to gravity (g = 9.8 m/s^2) and the time calculated earlier (t ≈ 5.07 s), we get:
v = 9.8 m/s^2 * 5.07 s

Calculating the value, the velocity (or speed) of the object just prior to splashing into the water is approximately 49.686 m/s.