# A Bullet Fired at an Angle of 60°: Time in the Air Calculation

## Determining the Time a Bullet Fired at 60° is in the Air

A bullet is fired at an angle of 60° to the horizontal with an initial velocity of 200 m/s. How long is the bullet in the air?

a) 40.8 s

b) 80.7 s

c) 50.4 s

d) 60.9 s

### Final answer:

The time a bullet fired at 60° with a velocity of 200 m/s is in the air is determined by its vertical motion. The total time of flight is 20.41 seconds, calculated using the projectile motion formula **t = (2 * v * sin(θ)) / g** where **v** is initial velocity, **θ** is launch angle, and **g** is gravitational acceleration.

### Explanation:

To determine how long a bullet fired at an angle of 60° to the horizontal with an initial velocity of 200 m/s is in the air, we must analyze the projectile motion of the bullet. Only the vertical motion affects the total time the bullet is in the air. Given this scenario, we will use the formula for the time of flight for a projectile launched at an angle:

**t = (2 * v * sin(θ)) / g**

where:

**v**is the initial velocity**θ**is the launch angle**g**is the acceleration due to gravity (9.81 m/s²)

In this case, plugging in the values gives:

**t = (2 * 200 m/s * sin(60°)) / 9.81 m/s²**

Calculating the sine of 60° which is √3/2 and inserting this with the other numbers into the formula, we can find the time the bullet is in the air. Therefore, **t = 20.41 s**, which represents **the time for projectile motion**.

**t = (2 * v * sin(θ)) / g**, where

**v**is the initial velocity,

**θ**is the launch angle, and

**g**is the acceleration due to gravity.