A bullet is fired with an initial velocity of 100 m/s at an angle of 30° from the level ground. Determine the amount of time the bullet spends in the air.
time to get to the top of the arc is when
50-9.8t = 0
now double that to get back to the ground.
because the initial velocity up is 100 sin 30 = 50 m/s
and the horizontal problem is irrelevant unless they ask for the range next.
To determine the amount of time the bullet spends in the air, we can use the laws of projectile motion. The motion of the bullet can be broken down into horizontal and vertical components.
The initial velocity of the bullet can be split into two components: the horizontal component (Vx) and the vertical component (Vy). To find the horizontal component, we use the formula:
Vx = V * cos(theta),
where V is the initial velocity of the bullet (100 m/s) and theta is the angle at which the bullet is fired (30°). Plugging in the values, we have:
Vx = 100 * cos(30°) = 100 * 0.866 = 86.6 m/s.
To find the vertical component, we use the formula:
Vy = V * sin(theta).
Plugging in the values, we have:
Vy = 100 * sin(30°) = 100 * 0.5 = 50 m/s.
Since we are only concerned with the vertical motion, we can disregard the horizontal motion for now.
The time taken for an object to reach its maximum height and come back down to the same level is the same. This is called the time of flight.
The time of flight can be found using the formula:
Time of flight = (2 * Vy) / g,
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we have:
Time of flight = (2 * 50) / 9.8 = 100 / 9.8 ≈ 10.20 seconds.
Therefore, the bullet spends approximately 10.20 seconds in the air.