A rock is dropped from a height of 2.7 m. How fast is it going when it hits the ground?
SUS
To find the speed of the rock when it hits the ground, we can use the equation for the final velocity of an object in free fall:
vf = sqrt(2 * g * h)
Where:
vf is the final velocity of the object
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height from which the object is dropped
Let's plug in the given values:
vf = sqrt(2 * 9.8 * 2.7)
Simplifying the equation:
vf = sqrt(52.92)
Calculating the square root:
vf ≈ 7.27 m/s
Therefore, the rock is going approximately 7.27 m/s when it hits the ground.
To find the speed at which the rock hits the ground, you can use the laws of motion and consider the effects of gravity.
The initial height of the rock is given as 2.7 meters, which means that it falls freely under the influence of gravity. The acceleration due to gravity (g) near the surface of the Earth is approximately 9.8 m/s^2.
We can use the equations of motion to determine the final speed of the rock when it hits the ground. In this case, we can use the equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (unknown)
- u is the initial velocity (0 m/s since it is dropped)
- a is the acceleration due to gravity (-9.8 m/s^2 for a falling object)
- s is the displacement (2.7 m)
Substituting the values, we get:
v^2 = 0^2 + 2 * (-9.8) * 2.7
v^2 = 0 + (-52.92)
v^2 = -52.92
The result is negative because we have considered the downward direction as positive. However, the magnitude of velocity or speed cannot be negative.
To find the magnitude of the velocity, we take the square root of both sides of the equation:
v = √(-52.92)
v ≈ √(-53)
Since we cannot take the square root of a negative number in real numbers, this indicates an error in our calculation. It suggests that our assumption of the downward direction as positive is incorrect, as the velocity should be negative.
By changing our assumption and treating the upward direction as positive, we can redo the calculation:
v^2 = 0^2 + 2 * (-9.8) * 2.7
v^2 = 0 + (-52.92)
v^2 = -52.92
Taking the square root of both sides:
v = √(-52.92)
v ≈ √(52.92)
v ≈ 7.27
Therefore, the speed at which the rock hits the ground is approximately 7.27 m/s.