An object dropped from a height h attain a velocity of 6m/s just before hitting the ground, find the value of h?

The equation to calculate the height h is as follows:

h = (1/2) * g * t^2 + v0 * t

Where g is the gravitational acceleration (9.8 m/s^2), t is the time for the object to fall, and v0 is the initial velocity of the object (0 m/s).

So, we can rearrange the equation to solve for t:

t = (-v0 + sqrt(v0^2 + 2 * g * h)) / g

Substituting in the values given, we get:

t = (-0 + sqrt(0^2 + 2 * 9.8 * h)) / 9.8

t = sqrt(2 * 9.8 * h) / 9.8

Now, we know that the object's final velocity is 6 m/s. So, we can use the equation:

v = v0 + g * t

Where v is the final velocity and v0 is the initial velocity (0 m/s).

Substituting in the values given, we get:

6 = 0 + 9.8 * t

6 = 9.8 * t

t = 6 / 9.8

Now, we can substitute this value of t back into the original equation to solve for h:

h = (1/2) * 9.8 * (6 / 9.8)^2 + 0 * (6/ 9.8)

h = (1/2) * 9.8 * 36 / 96 + 0

h = 18 / 48

h = 0.375 m