To find the time it takes for the rock to hit the ground and its velocity at that moment, we can use the equations of motion under constant acceleration, assuming the only force acting on the rock is gravity.
Let's consider the downward direction as positive (taking into account the conventional coordinate system where the positive y-direction is upward).
Given: Initial height (h₀) = 6 m Acceleration due to gravity (g) = 9.8 m/s²
To find the time it takes for the rock to hit the ground, we can use the equation:
h = h₀ + v₀t + (1/2)gt²
Since the rock is dropped from rest, the initial velocity (v₀) is zero. Therefore, the equation simplifies to:
h = h₀ + (1/2)gt²
Plugging in the values:
6 = 0 + (1/2) * 9.8 * t²
Rearranging the equation:
9.8t² = 12
t² = 12 / 9.8
t² ≈ 1.224
Taking the square root of both sides:
t ≈ √1.224
t ≈ 1.105 seconds
So, it takes approximately 1.105 seconds for the rock to hit the ground.
To find the velocity of the rock as it hits the ground, we can use the equation:
v = v₀ + gt
Since the initial velocity is zero, the equation becomes:
v = gt
Plugging in the values:
v = 9.8 * 1.105
v ≈ 10.822 m/s
Therefore, the velocity of the rock as it hits the ground is approximately 10.822 m/s, directed downward.