A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 32 m/s .

To find the initial speed of the rock, we can use the kinematic equation for free-fall motion. The equation we can use is:

v^2 = u^2 + 2as

Where:
v = final velocity (32 m/s)
u = initial velocity (what we're trying to find)
a = acceleration due to gravity (-9.8 m/s^2, because it's acting opposite to the motion)
s = displacement (32 m, because the rock has fallen from the top of the cliff to the ground)

Part A:
- Rearranging the equation, we get: u^2 = v^2 - 2as
- Plugging in the given values: u^2 = (32 m/s)^2 - 2(-9.8 m/s^2)(32 m)
u^2 = 1024 m^2/s^2 + 627.2 m^2/s^2
u^2 = 1651.2 m^2/s^2
- Taking the square root of both sides, we get: u = √1651.2 m/s ≈ 40.64 m/s

Therefore, the initial speed of the rock is approximately 40.64 m/s.

Part B:
To find the greatest height of the rock, we can use the equation for the vertical motion of an object in free fall:

v^2 = u^2 + 2as

However, in this case, we are interested in finding the maximum height reached by the rock, so the final velocity (v) at the highest point will be zero.

0 = u^2 + 2as

The acceleration (a) is still -9.8 m/s^2 and the displacement (s) is what we need to find.

- Rearranging the equation, we get: s = -u^2 / (2a)
- Plugging in the known values: s = -(40.64 m/s)^2 / (2 * -9.8 m/s^2)
s = -1650.5 m^2/s^2 / (-19.6 m/s^2)
s ≈ 84.28 m

Therefore, the greatest height of the rock, as measured from the base of the cliff, is approximately 84.28 m.