A pistol is fired horizontally and the bullet has a muzzle velocity (speed with which the bullet leaves the barrel) of 31.3 m/s. The gun is 50 m above the surrounding terrain.
a) How far away does the bullet land measured horizontally from a point directly beneath the gun? You can see the exact situation at this site:
b) The gun is now tilted upward at an angle of 45 degrees and fired with the same muzzle velocity. What is its horizontal range now? (Again you can check your answer on the above website). You should refresh yourself on the solution to the Quadratic Equation.
A pistol is fired horizontally and the bullet has a muzzle velocity (speed with which the bullet leaves the barrel) of 31.3 m/s. The gun is 50 m above the surrounding terrain. The gun is tilted upward at an angle of 45 degrees.
What is its horizontal range now?
a) To find the distance that the bullet lands horizontally from a point directly beneath the gun, we need to consider the horizontal motion of the bullet.
Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion. The time taken for the bullet to land can be found using the equation:
Distance = Velocity × Time
The initial velocity in the horizontal direction is the same as the muzzle velocity of the bullet, which is 31.3 m/s.
The vertical motion of the bullet can be described using the equation of motion:
Vertical Distance = Vertical Initial Velocity × Time + (0.5 × Acceleration × Time^2)
Since the bullet is fired horizontally, the vertical component of the initial velocity is 0.
The acceleration due to gravity, denoted by 'g', is approximately equal to 9.8 m/s^2.
The vertical distance traveled by the bullet is equal to the height of the gun, which is 50 m.
We can rearrange the equation of motion to solve for time:
Vertical Distance = (0.5 × Acceleration × Time^2)
50 m = 0.5 × 9.8 m/s^2 × Time^2
Simplifying this equation, we get:
Time^2 = 50 m × 2 / 9.8 m/s^2
Time^2 = 10.2 s^2
Time = √(10.2 s^2) ≈ 3.19 s
Now, we can calculate the horizontal distance traveled by the bullet:
Distance = Velocity × Time
Distance = 31.3 m/s × 3.19 s
Distance ≈ 99.95 m
Therefore, the bullet lands approximately 99.95 meters away horizontally from a point directly beneath the gun.
b) When the gun is tilted upward at an angle of 45 degrees, the horizontal range is different because now there is an initial horizontal velocity and an initial vertical velocity.
The horizontal component of the initial velocity can be found using the muzzle velocity and the angle of 45 degrees:
Horizontal Initial Velocity = Muzzle Velocity × cos(45 degrees)
Horizontal Initial Velocity = 31.3 m/s × cos(45 degrees)
Horizontal Initial Velocity ≈ 22.17 m/s
The vertical component of the initial velocity can also be found using the muzzle velocity and the angle of 45 degrees:
Vertical Initial Velocity = Muzzle Velocity × sin(45 degrees)
Vertical Initial Velocity = 31.3 m/s × sin(45 degrees)
Vertical Initial Velocity ≈ 22.17 m/s
The time taken for the bullet to land can be found using the vertical motion equation as before. The vertical distance traveled by the bullet is still equal to the height of the gun, which is 50 m.
Using the equation of motion:
50 m = (Vertical Initial Velocity × Time) + (0.5 × Acceleration × Time^2)
Substituting the values:
50 m = (22.17 m/s × Time) + (0.5 × 9.8 m/s^2 × Time^2)
Rearranging the equation, we get:
0.5 × 9.8 m/s^2 × Time^2 + 22.17 m/s × Time - 50 m = 0
This is a quadratic equation of the form: ax^2 + bx + c = 0, where a = 0.5 × 9.8 m/s^2, b = 22.17 m/s, and c = -50 m.
We can solve this quadratic equation using the quadratic formula:
Time = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
Time = (-22.17 m/s ± √((22.17 m/s)^2 - 4 × 0.5 × 9.8 m/s^2 × (-50 m))) / (2 × 0.5 × 9.8 m/s^2)
Simplifying this equation, we find two possible values for the time:
Time ≈ 3.80 s or Time ≈ 0.60 s
Since we are looking for the time the bullet takes to land, we choose the positive value:
Time ≈ 3.80 s
Now, we can calculate the horizontal range:
Horizontal Range = Horizontal Initial Velocity × Time
Horizontal Range = 22.17 m/s × 3.80 s
Horizontal Range ≈ 84.11 m
Therefore, the bullet lands approximately 84.11 meters away horizontally when the gun is tilted upward at an angle of 45 degrees.
To find the horizontal distance the bullet lands in both scenarios, we need to use the equations of projectile motion.
a) When the pistol is fired horizontally, the bullet will have an initial horizontal velocity (Vx) of 31.3 m/s and an initial vertical velocity (Vy) of zero. The only force acting on the bullet is gravity, which causes it to accelerate downward at 9.8 m/s^2.
We can use the equation for horizontal distance (d):
d = Vx * t
Since there is no initial vertical velocity, the time taken to fall to the ground can be calculated using the equation for vertical displacement (d):
d = (1/2) * g * t^2
where g is the acceleration due to gravity (9.8 m/s^2). Rearranging the equation gives:
t = sqrt((2 * d) / g)
Substituting the given values, we have:
t = sqrt((2 * 50) / 9.8) = 3.19 s
Now, we can substitute the value of t back into the equation for horizontal distance:
d = 31.3 m/s * 3.19 s = 99.7 m
So, the bullet will land approximately 99.7 meters away measured horizontally from a point directly beneath the gun.
b) When the gun is tilted upward at an angle of 45 degrees, the initial velocity of the bullet can be split into horizontal (Vx) and vertical (Vy) components. The horizontal component remains the same at 31.3 m/s, while the vertical component can be calculated using the angle and the initial velocity.
Vy = V * sin(θ)
where V is the initial velocity (31.3 m/s) and θ is the angle of elevation (45 degrees).
Vy = 31.3 m/s * sin(45 degrees) = 22.1 m/s
To find the horizontal range in this scenario, we again need to find the time of flight. The time taken for the bullet to reach its maximum height can be calculated using the equation for vertical velocity:
Vy = g * t_max
Rearranging the equation gives:
t_max = Vy / g
t_max = 22.1 m/s / 9.8 m/s^2 = 2.25 s
Since the bullet reaches its maximum height halfway into its total time of flight, the total time of flight (T) is twice the time to reach the maximum height:
T = 2 * t_max = 2 * 2.25 s = 4.5 s
Now, we can substitute the value of T back into the equation for horizontal distance:
d = Vx * T
d = 31.3 m/s * 4.5 s = 140.85 m
So, when the gun is fired at an angle of 45 degrees, the bullet will land approximately 140.85 meters away measured horizontally from a point directly beneath the gun.