an ironworker drops a hammer 5.25m to the ground. what is its speed as it hits the ground?
V = sqrt(2 g H)
H = 5.25 m
g = 9.8 m/s^2
Do the numbers
To find the speed of the hammer as it hits the ground, we can use the equations of motion. We will assume that the only force acting on the hammer is gravity.
The equation that relates distance, initial velocity, final velocity, and acceleration is:
(v_f)^2 = (v_i)^2 + 2 * a * d
Where:
- (v_f) is the final velocity (the velocity of the hammer as it hits the ground),
- (v_i) is the initial velocity (which is zero because the hammer is dropped),
- a is the acceleration due to gravity (approximately 9.8 m/s^2),
- d is the distance traveled (which is 5.25 m).
Now, let's plug in the values into the equation and solve for (v_f):
(v_f)^2 = (0)^2 + 2 * 9.8 * 5.25
(v_f)^2 = 0 + 2 * 9.8 * 5.25
(v_f)^2 = 102.9
Taking the square root of both sides:
v_f = √102.9
v_f ≈ 10.14 m/s
Therefore, the speed of the hammer as it hits the ground is approximately 10.14 m/s.
To calculate the speed at which the hammer hits the ground, we can use the equation for the final velocity of an object in free fall:
v = sqrt(2 * g * h)
where:
v is the final velocity (speed) of the hammer
g is the acceleration due to gravity (approximately 9.8 m/s²)
h is the height from which the hammer is dropped (5.25 m in this case)
Now let's calculate the speed of the hammer as it hits the ground:
v = sqrt(2 * 9.8 m/s² * 5.25 m)
v = sqrt(102.9 m²/s²)
v ≈ 10.1 m/s
Therefore, the speed at which the hammer hits the ground is approximately 10.1 m/s.