A worker drops a wrench from the top of a tower 118.5 m tall. What is the velocity when the wrench strikes the ground?

m g h = 1/2 m v^2 ... v = √(2 g h)

confused

To find the velocity when the wrench strikes the ground, we can use the basic equations of motion. First, we need to determine the time it takes for the wrench to fall.

We can use the equation for displacement in vertical motion:

s = ut + (1/2)gt^2,

where s is the displacement (118.5 m), u is the initial velocity (which is zero because the wrench is dropped), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

Substituting the given values into the equation, we get:

118.5 = 0 × t + (1/2) × (-9.8) × t^2.

Simplifying this equation, we have:

118.5 = -4.9 t^2.

Rearranging the equation, we get:

t^2 = 118.5 / -4.9.

Taking the square root of both sides, we find:

t ≈ √(118.5 / -4.9).

Evaluating this expression, we get:

t ≈ -√(-24.18).

Since time cannot be negative in this context, we ignore the negative sign and consider the positive value:

t ≈ √(24.18).

Therefore, the time it takes for the wrench to fall is approximately √(24.18) seconds.

Now we can find the final velocity using the equation:

v = u + gt,

where v is the final velocity, u is the initial velocity (zero in this case), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time (√(24.18) seconds).

Substituting the given values into the equation, we get:

v = 0 + (-9.8) × √(24.18).

Evaluating this expression, we find:

v ≈ -47.68.

Since velocity cannot be negative in this context, we consider the magnitude of the velocity:

v ≈ 47.68 m/s.

Therefore, the velocity when the wrench strikes the ground is approximately 47.68 m/s.