# How to Calculate How Long a Projectile Is Airborne

When you launch a projectile into the air, you can use physics to determine how long it will remain airborne. Because the force of gravity only acts downward — that is, in the vertical direction — you can treat the vertical and horizontal components of the flight path separately. As a result, you can calculate things like the time the projectile will be airborne before it strikes the ground.

Say, for example, that you decide to shoot a cannon into the air. Assume that it has a muzzle velocity of 860 meters/second, and that it is pointing straight up. How long do you have to get out of the way before the cannonball comes back down and obliterates your new cannon?

First, you have to determine how long it will take for the cannonball to reach its maximum height. You know that the vertical velocity of the cannonball at its maximum height is 0 meters/second, so you can use the following equation to find the time the cannonball will take to reach its maximum height:

*v** _{f}* =

*v*

*+*

_{i}*at*

Because *v** _{f}* = 0 meters/second and

*a*= –

*g*= –9.8 meters/seconds

^{2}, it works out to this:

0 = *v** _{i}* –

*gt*

Solving for time, you get the following:

You enter the numbers into your calculator as follows:

It takes about 88 seconds for the cannonball to reach its maximum height (ignoring air resistance).

So how long will it take the cannonball to complete its entire trip? Flights like the one taken by the cannonball are symmetrical; the trip up is a mirror of the trip down. The velocity at any point on the way up has exactly the same magnitude as on the way down, but on the way down, the velocity is in the opposite direction. This means that the total flight time is double the time it takes the cannonball to reach its highest point, or

*t** _{total}* = 2(88 s) = 176 s

You have 176 seconds, or 2 minutes and 56 seconds, until the cannonball destroys the cannon that fired it.