Abody is released from a height of 20m above the ground. Find it's height above the ground when it's velocity is 10m/s

vf^2=vi^2 + 2gdx where x is falling height, so if d is height, the falling height is 20-d

100=0+2(-9.8)*(20-d)
100/19.6-20=-d
d=you do it, about 15m above ground

To find the height above the ground when the velocity is 10 m/s, you can use the equations of motion and the principles of kinematics.

Step 1: Understand the variables and equations involved:
- The height above the ground is represented by "h."
- The initial velocity is zero because the body is released from rest at a height of 20m.
- The final velocity is 10 m/s.
- The acceleration due to gravity is constant and equal to -9.8 m/s^2 (taking downward direction as negative).

Step 2: Identify the equation that relates the variables:
We can use the equation of motion that relates the final velocity, initial velocity, acceleration, and displacement in the vertical direction. The equation is:
v_f^2 = v_i^2 + 2aΔh

Step 3: Rearrange the equation and solve for the unknown variable:
Replacing the known values in the equation, we get:
10^2 = 0^2 + 2(-9.8)Δh

Simplifying further:
100 = -19.6Δh

Divide both sides of the equation by -19.6 to solve for Δh:
Δh = -100 / -19.6 ≈ 5.1 m

Therefore, the height above the ground when the velocity is 10 m/s is approximately 5.1 meters.