A rifle is aimed horizontally toward the center of the target 220 m away. If the bullet strikes 48 cm below the center, what was the velocity of the bullet (ignore air friction)?
To determine the velocity of the bullet, we can use the principle of projectile motion. Assuming the rifle is initially at the same height as the center of the target, and neglecting air friction, we can use the equations of motion:
1. The horizontal distance traveled by the bullet can be calculated using the formula:
range = velocity × time
Since the rifle is aimed horizontally, the time of flight for the bullet is the same as the time for the bullet to hit the target.
2. The vertical displacement of the bullet is given by:
vertical displacement = initial vertical velocity × time + (1/2) × acceleration due to gravity × time^2
Since the bullet is lower than the center of the target, the initial vertical velocity is downward.
We need to calculate the initial vertical velocity of the bullet.
Let's assume the downward direction is negative.
Given:
Range (horizontal distance) = 220 m
Vertical displacement = -0.48 m
Acceleration due to gravity (g) = 9.8 m/s^2
Step 1: Calculate the time of flight
Using the horizontal distance formula, rearranged for time:
range = velocity × time
220 = velocity × time
Solving for time:
time = 220 / velocity
Step 2: Calculate the initial vertical velocity
Using the formula for vertical displacement:
vertical displacement = initial vertical velocity × time + (1/2) × acceleration due to gravity × time^2
Plugging in the values:
-0.48 = initial vertical velocity × time - (1/2) × 9.8 × time^2
Substituting the expression for time from Step 1:
-0.48 = initial vertical velocity × (220 / velocity) - (1/2) × 9.8 × (220 / velocity)^2
Simplifying:
-0.48 = 220 × initial vertical velocity / velocity - 4.9 × 220^2 / velocity^2
Rearranging the equation:
0 = 220 × initial vertical velocity / velocity - 4.9 × 220^2 / velocity^2 + 0.48
Step 3: Solve the equation
This equation is quadratic in terms of velocity. We can use the quadratic formula to solve it:
velocity = [-b ± √(b^2 - 4ac)] / 2a
Where:
a = -4.9 × 220^2
b = 220 × initial vertical velocity
c = -0.48
Substituting the values:
velocity = [-220 × initial vertical velocity ± √((220 × initial vertical velocity)^2 + (4 × 4.9 × 220^2 × 0.48))] / (2 × -4.9 × 220^2)
This equation will give us the possible values for the velocity of the bullet.
To find the velocity of the bullet, we can use the principles of projectile motion.
First, let's assume that the bullet was in the air for time 't'. In this case, we can consider the horizontal and vertical components of the bullet's motion separately.
1. Vertical Motion:
The bullet strikes the target 48 cm below the center, which means it traveled a vertical distance of 0.48 meters (converting cm to meters). The initial vertical velocity is 0, as the bullet is aimed horizontally. The acceleration due to gravity, 'g', is approximately 9.8 m/s² (ignoring air resistance).
We can use the formula for vertical displacement in projectile motion:
δy = (v₀y * t) + (0.5 * g * t²)
where δy is the vertical displacement, v₀y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Plugging in the values, we have:
0.48 = (0 * t) + (0.5 * 9.8 * t²)
0.48 = 4.9 * t²
Now, let's solve for t:
t² = 0.48 / 4.9
t² ≈ 0.098
t ≈ √0.098
t ≈ 0.313 seconds
2. Horizontal Motion:
The horizontal distance traveled by the bullet is 220 meters, and there is no horizontal acceleration in this case. We can use the formula for horizontal displacement in projectile motion:
δx = v₀x * t
where δx is the horizontal displacement, v₀x is the initial horizontal velocity, and t is the time.
Plugging in the values, we have:
220 = v₀x * 0.313
v₀x = 220 / 0.313
v₀x ≈ 702.57 m/s
Therefore, the velocity of the bullet is approximately 702.57 m/s.