# A bullet is fired straight down from a 100 meter tower. The mass of the bullet is 5 grams and the velocity is 250 meters/second. What is the speed of the bullet just before it hits the ground?

Jul 4, 2015

The bullet will hit the ground with a velocity of 254 m/s.

#### Explanation:

So, you know that the bullet is being fired from a height of 100 m and with an initial velocity of 250 m/s.

In this particular instance, the actual mass of the bullet is not important. You can approach this problem by thinking about energy conservation.

The energy the bullet has at the top of the bulding must be equal to the energy it has when it hits the ground.

At the top of the building, the bullet has gravitational potential energy and kinetic energy

${E}_{\text{top}} = m \cdot g \cdot h + \frac{1}{2} \cdot m \cdot {v}_{0}^{2}$

When it makes contact with the ground, it will only have kinetic energy

${E}_{\text{ground}} = \frac{1}{2} \cdot m \cdot {v}_{f}^{2}$

This means that you have

${E}_{\text{top" = E_"ground}}$

$\cancel{m} \cdot g \cdot h + \frac{1}{2} \cancel{m} \cdot {v}_{0}^{2} = \frac{1}{2} \cancel{m} \cdot {v}_{f}^{2}$

This is equivalent to

${v}_{f}^{2} = {v}_{0}^{2} + 2 \cdot g \cdot h$

Plug in your values and solve for ${v}_{f}$ to get

${v}_{f} = \sqrt{{v}_{0}^{2} + 2 \cdot g \cdot h}$

${v}_{f} = \sqrt{250 {\text{^2"m"^2/"s"^2 + 2 * 10"m"/"s"^2 * 100"m") = sqrt(64500"m"^2/"s}}^{2}}$

${v}_{f} = \textcolor{g r e e n}{\text{254 m/s}}$