# A bullet is shot from the East Coast of the USA and reaches the West Coast of the USA in 4.5 seconds. How fast is that in kilometers per hour?

4.5 seconds; $4 , 173 , 120 \frac{k m}{h r}$

#### Explanation:

I'm going to tackle this, with a couple of assumptions:

1. There is no wind resistance, and
2. There is no bullet deflection (aside from what is needed to deal with the curvature of the Earth - 3240 miles being roughly the distance between the East and West coasts of the USA.

To solve for time in the air, we use:

$s = v t$

We know $s$ and $v$ so let's solve for $t$:

$3240 = 720 t$

$t = 4.5 \sec$

The velocity of the bullet in $\text{km"/"hr}$ can be found using unit conversions:

$720 \frac{\cancel{m i}}{\cancel{\sec}} \left(\frac{3600 \cancel{\sec}}{1 h r}\right) \left(\frac{1.61 k m}{1 \cancel{m i}}\right) = 4 , 173 , 120 \frac{k m}{h r}$

Just for fun, let's figure out how far that is.

The ISS (International Space Station) is at an altitude of 6371 + 400 = 6771 km from the centre of the Earth (the 6371 is the radius of the Earth and the 400 is the ISS altitude from the surface). Using the circumference of a circle:

$C = 2 \pi r$, we can see that the ISS traverses:

$C = 2 \pi \left(6771\right) = 42543 k m$ and takes roughly an hour and a half (92 minutes).

Our bullet would travel roughly 6,260,000 km in an hour and a half, meaning in the time the ISS orbited the Earth once, the bullet would do so close to 150 times.