A rifle is aimed horizontally toward the center of a target 220 m away. If the bullet strikes 48 cm below the center, what was the velocity of the bullet (ignore the friction)?
To find the velocity of the bullet, we can use the kinematic equation for vertical motion:
d = (V^2 * sin^2θ) / (2 * g)
Where:
d is the vertical displacement of the bullet (given as 48 cm = 0.48 m)
V is the initial velocity of the bullet
θ is the angle of launch (0 degrees since the rifle is aimed horizontally)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the rifle is aimed horizontally, the initial velocity of the bullet is the horizontal velocity of the bullet.
The horizontal distance traveled by the bullet is 220 m. So, the time of flight can be found using the formula d = Vx * t, where Vx is the horizontal component of the velocity and t is the time of flight.
220 m = Vx * t (Equation 1)
Now, let's solve for t.
Since the vertical displacement is given as 48 cm below the center, the time of flight can also be determined using the formula d = (1/2) * g * t^2, where t is the time of flight.
0.48 m = 0.5 * 9.8 m/s^2 * t^2 (Equation 2)
We can solve Equation 2 for t:
0.48 m = 4.9 m/s^2 * t^2
t^2 = 0.48 m / 4.9 m/s^2
t^2 ≈ 0.0979 s^2
t ≈ √(0.0979 s^2)
t ≈ 0.3129 s
Now, substitute the value of t into Equation 1 to find Vx:
220 m = Vx * 0.3129 s
Vx = 220 m / 0.3129 s
Vx ≈ 702.27 m/s
Therefore, the velocity of the bullet (horizontal velocity) is approximately 702.27 m/s.
To find the velocity of the bullet, we can use the equation for horizontal projectile motion. We know that the horizontal distance traveled by the bullet is 220 m, and the vertical distance below the center of the target is 48 cm.
Given:
Horizontal distance (range) (x) = 220 m
Vertical distance (y) = -48 cm = -0.48 m (since it is below the center)
Now, let's break down the problem into horizontal and vertical components:
The horizontal component of velocity (vx) remains constant throughout the motion because there is no horizontal acceleration (assuming no air resistance).
The vertical component of velocity (vy) changes over time due to the acceleration due to gravity.
Now, we need to find the time of flight (t) and the initial vertical velocity (viy) of the bullet.
First, let's find the time of flight (t):
Since the only acceleration acting on the bullet is due to gravity, we can use the equation y = viy * t + (1/2) * g * t^2, where viy is the initial vertical velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In our case, at the highest point of the projectile's trajectory, the vertical displacement (y) will be zero (because it will be at the same height as the center of the target). So, we can rewrite the equation as 0 = viy * t + (1/2) * g * t^2.
Since the bullet was aimed horizontally, the initial vertical velocity (viy) is 0 (because there is no initial vertical velocity).
0 = 0 * t + (1/2) * 9.8 * t^2
Simplifying the equation:
0 = (1/2) * 9.8 * t^2
Solving for t:
0 = 4.9 * t^2
t^2 = 0 (since the only solution for this case is when t = 0, which corresponds to the bullet hitting the target)
Therefore, the time of flight (t) is 0 seconds.
Now, let's find the initial horizontal velocity (vix) of the bullet:
We can use the equation x = vix * t, where x is the horizontal distance (range) and t is the time of flight.
220 = vix * 0 (since t = 0)
Therefore, the initial horizontal velocity (vix) is undefined because the time of flight is 0 seconds.
To summarize, since the time of flight is 0 seconds, the bullet would have needed an infinite initial velocity in order to hit the target.