A hammer falls from a scaffold on a building 50.0 m above the ground. Find its speed as it hits the
ground.
1/2 mv^2=mg(50)
v=sqrt(100*9.8)
To find the speed of the hammer as it hits the ground, we can use the principles of physics, specifically the equation of motion for objects falling under the influence of gravity.
The initial velocity of the hammer is zero since it falls freely from rest. The only force acting on the hammer is gravity, which causes it to accelerate downwards.
We can use the following equation to find the speed of the hammer as it hits the ground:
v^2 = u^2 + 2as
Where:
- v is the final velocity (speed)
- u is the initial velocity (which is 0 in this case)
- a is the acceleration due to gravity (approximately 9.8 m/s^2)
- s is the distance traveled (which is 50.0 m in this case)
Plugging in the values into the equation, we get:
v^2 = 0^2 + 2 * 9.8 * 50.0
v^2 = 0 + 2 * 9.8 * 50.0
v^2 = 980
Now, taking the square root of both sides of the equation, we get:
v = sqrt(980)
v ≈ 31.3 m/s
Therefore, the speed of the hammer as it hits the ground is approximately 31.3 m/s.