a gun tilted upward at 30 degrees fires a bullet with a speed of 215 m/s. what is the bullets velocity parallel to the ground
Well, let me take a shot at this question! When a gun is tilted upward at any angle, the bullet's velocity can be broken down into two components: one parallel to the ground and one perpendicular to the ground. In this case, the gun is tilted upward at 30 degrees and the bullet speed is 215 m/s.
Now, to find the bullet's velocity parallel to the ground, we simply need to determine the horizontal component of its velocity. We can do this by multiplying the bullet's speed by the cosine of the angle. So, let's crunch some numbers!
Velocity parallel to the ground = 215 m/s * cos(30 degrees)
And using some hilarious math skills, we find that the bullet's velocity parallel to the ground is approximately 186.34 m/s.
So, watch out, world! This bullet is definitely on the move, but thankfully it's parallel to the ground, which means it won't fly away into space!
To find the velocity of the bullet parallel to the ground, we need to determine the horizontal component of the initial velocity. The angle of the gun being tilted upward at 30 degrees is considered as the launch angle.
The horizontal component of the initial velocity can be determined using the formula:
Horizontal velocity = Initial velocity * Cos(angle)
Given:
Initial velocity (v) = 215 m/s
Angle (θ) = 30 degrees
Substituting the given values into the formula:
Horizontal velocity = 215 m/s * Cos(30 degrees)
We need to convert the angle from degrees to radians since the trigonometric functions in most programming languages accept radians as input.
Using the conversion: Radians = Degrees * (π/180)
Where π is approximately 3.14159.
θ in radians = 30 degrees * (π/180)
θ in radians = 0.5236 radians
Now, substituting the values into the formula:
Horizontal velocity = 215 m/s * Cos(0.5236 radians)
Evaluating the equation:
Horizontal velocity = 215 m/s * 0.866
Horizontal velocity = 186.19 m/s
Therefore, the bullet's velocity parallel to the ground is approximately 186.19 m/s.
To find the bullet's velocity parallel to the ground, we need to break down the initial velocity into its vertical and horizontal components.
The vertical component of the initial velocity can be found using the equation:
Vvertical = Vinitial * sin(angle)
Where:
Vvertical is the vertical component of the initial velocity
Vinitial is the initial velocity of the bullet, which is given as 215 m/s
angle is the tilt angle of the gun, which is given as 30 degrees
Substituting the values into the equation:
Vvertical = 215 * sin(30º)
Vvertical ≈ 107.5 m/s
The horizontal component of the initial velocity can be found using the equation:
Vhorizontal = Vinitial * cos(angle)
Where:
Vhorizontal is the horizontal component of the initial velocity
Vinitial is the initial velocity of the bullet, which is given as 215 m/s
angle is the tilt angle of the gun, which is given as 30 degrees
Substituting the values into the equation:
Vhorizontal = 215 * cos(30º)
Vhorizontal ≈ 186.4 m/s
Since the bullet's vertical velocity doesn't affect its motion parallel to the ground, the bullet's velocity parallel to the ground is equal to the horizontal component of its initial velocity.
Therefore, the bullet's velocity parallel to the ground is approximately 186.4 m/s.
Initial velocity was 215 m/s at θ=30° upwards of the horizontal.
The horizontal component of the velocity is its projection on the horizontal, namely: magnitude = 215cos(θ) and direction: along the horizontal.