To determine the speed of the object when it hits the ground, we can use the principle of conservation of mechanical energy. The object's potential energy at the height of 6 meters will be completely converted into kinetic energy just before it hits the ground.

The relevant formulas are:

1. Potential energy (PE):
   $PE = mgh$ where:

   • $m$ is the mass of the object (13 kg),
   • $g$ is the acceleration due to gravity (9.8 m/s²),
   • $h$ is the height (6 m).

2. Kinetic energy (KE):
   $KE = \frac{1}{2}mv^2$ where:

   • $v$ is the velocity of the object when it hits the ground.

Since energy is conserved, the potential energy at the height will equal the kinetic energy just before impact: $mgh = \frac{1}{2}mv^2$

Now we can solve for $v$: $v = \sqrt{2gh}$

Substituting in the known values: $v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 6 \, \text{m}}$ $v = \sqrt{117.6 \, \text{m}^2/\text{s}^2}$ $v \approx 10.84 \, \text{m/s}$

Thus, the object will be moving at approximately 10.84 m/s when it hits the ground.

Would you like a more detailed breakdown or have any other questions?

Related Questions:

1. How would the speed change if the object were dropped from a higher height?
2. What would happen if air resistance were considered in this problem?
3. How can we calculate the time it takes for the object to reach the ground?
4. How does the mass of the object affect its speed upon hitting the ground?
5. What is the kinetic energy of the object just before it hits the ground?

Tip:

In problems involving free fall, the mass of the object doesn't affect its final velocity—only the height and gravitational acceleration matter, as long as air resistance is negligible.