A rifle is fired horizontally. The bullet is found to have traveled 200 meters. If the rifle barrel is 1.90 meters above the ground, at what speed must the bullet have been traveling as it left the barrel?
56*7
52
To find the speed at which the bullet must have been traveling as it left the barrel, we can use the formula:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s as it starts from rest)
a = acceleration (0 m/s^2 horizontally)
s = displacement (200 m horizontally)
Since there is no acceleration in the horizontal direction, we can simplify the equation to:
v^2 = 0^2 + 2 * 0 * 200
v^2 = 0 + 0
v^2 = 0
Therefore, the speed at which the bullet must have been traveling as it left the barrel is 0 m/s.
To determine the speed at which the bullet must have been traveling as it left the barrel, we can use the principles of projectile motion.
1. First, let's define the relevant variables:
- Initial velocity (u) of the bullet
- Distance traveled horizontally (d) = 200 meters
- Height above the ground (h) = 1.90 meters
- Acceleration due to gravity (g) = approximately 9.8 m/s² (assuming no air resistance)
2. The time taken (t) for the bullet to travel the horizontal distance is the same as the time taken for it to fall vertically to the ground. Since there is no initial vertical velocity, we can use the kinematic equation:
h = (1/2) * g * t²
Rearranging the equation, we have:
t = sqrt((2h) / g)
Substituting the values:
t = sqrt((2 * 1.90) / 9.8)
t ≈ 0.616 seconds (rounded to three decimal places)
3. Now, we can calculate the initial horizontal velocity (u) using the formula:
u = d / t
Substituting the values:
u = 200 / 0.616
u ≈ 324.68 m/s (rounded to two decimal places)
Therefore, the bullet must have been traveling at approximately 324.68 m/s (or meters per second) as it left the barrel.