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 .
Part A
Assuming that air resistance can be ignored, find the initial speed of the rock.
Part B
Find the greatest height of the rock as measured from the base of the cliff.
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.