To determine the time it takes for the rock to hit the ground and the velocity at impact, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to gravity and is approximately 9.8 m/s² (assuming no significant air resistance).
First, let's calculate the time it takes for the rock to fall to the ground. We can use the equation:
s = ut + (1/2)at²
Where: s is the vertical distance (6 meters in this case) u is the initial velocity (which is 0 since the rock is dropped) a is the acceleration due to gravity (-9.8 m/s²) t is the time we want to find
Plugging in the values:
6 = 0 * t + (1/2)(-9.8)t²
Simplifying:
6 = -4.9t²
Dividing both sides by -4.9:
t² = -6 / -4.9
t² ≈ 1.2245
Taking the square root of both sides:
t ≈ √1.2245
t ≈ 1.105 seconds
Therefore, it takes approximately 1.105 seconds for the rock to hit the ground.
Next, let's calculate the velocity of the rock as it hits the ground. We can use the equation:
v = u + at
Where: v is the final velocity (which is what we want to find) u is the initial velocity (0 m/s since the rock is dropped) a is the acceleration due to gravity (-9.8 m/s²) t is the time it takes to hit the ground (1.105 seconds)
Plugging in the values:
v = 0 + (-9.8)(1.105)
v ≈ -10.824 m/s
Note that the negative sign indicates that the velocity is downward, which is consistent with the direction of motion as the rock falls.
Therefore, the velocity of the rock as it hits the ground is approximately -10.824 m/s.