A worker drops a wrench from the top of a tower 95.1 m tall. What is the velocity when the wrench strikes the ground?

h=v²/2g,

v=sqrt(2gh)=sqrt(2•9.8•95.1) =43.17 m/s

since velocity of any falling object is the gravitational pull, it is 9.8 m/s. *the negative or positive of the answer depends on the positive direction.

To find the velocity of the wrench when it strikes the ground, we can use the equation of motion for objects in freefall:

v^2 = u^2 + 2as

where:
- v is the final velocity (the velocity when the wrench strikes the ground)
- u is the initial velocity (which is 0 since the wrench is dropped from rest)
- a is the acceleration due to gravity (which is approximately 9.8 m/s^2)
- s is the distance covered (which is the height of the tower, 95.1 m)

Let's substitute the given values into the equation and solve for v:

v^2 = 0^2 + 2(9.8)(95.1)

v^2 = 0 + 1861.96

v^2 = 1861.96

Taking the square root of both sides:

v = √(1861.96)

v ≈ 43.13 m/s

Therefore, the velocity of the wrench when it strikes the ground is approximately 43.13 m/s.

To find the velocity of the wrench when it strikes the ground, we can use the principles of physics related to free fall. The key formula we need is the kinematic equation for final velocity in free fall:

v = √(2gh)

where v is the final velocity, g is the acceleration due to gravity, and h is the height.

First, we need to determine the value of g, which is approximately 9.8 m/s² on Earth.

Next, we can substitute the given values into the formula:

v = √(2 * 9.8 * 95.1)

Now, let's calculate it:

v = √(1869.6)

v ≈ 43.24 m/s

Therefore, the velocity of the wrench when it strikes the ground is approximately 43.24 m/s.