An 10g bullet leaves a gun at 800 m/s. What is the maximum distance that the bullet could reach? What is the height it reaches when the angle above the horizontal is 45 degrees

Question

An 10g bullet leaves a gun at 800 m/s. What is the maximum distance that the bullet could reach? What is the height it reaches when the angle above the horizontal is 45 degrees

Answer 1

To find the maximum distance that the bullet could reach, we can use the equations of motion for projectile motion. The key assumption here is that there is no air resistance.

First, we need to find the time it takes for the bullet to reach its maximum height. In projectile motion, the time taken to reach the highest point of the trajectory is half of the total time of flight.

The time of flight (T) can be calculated using the equation:

T = (2 * initial vertical velocity) / gravity

Where the initial vertical velocity is the vertical component of the initial velocity and gravity is the acceleration due to gravity (9.8 m/s^2).

Since the bullet is fired at an angle of 45 degrees above the horizontal, we can calculate the initial vertical velocity (V_y) using the equation:

V_y = initial velocity * sin(theta)

Where theta is the angle above the horizontal (in this case, 45 degrees).

Using these values, we can calculate the time of flight:

V_y = 800 m/s * sin(45 degrees)
≈ 800 m/s * 0.7071
≈ 565.68 m/s

T = (2 * 565.68 m/s) / 9.8 m/s^2
≈ 115.32 s

Now that we have the time of flight, we can calculate the maximum height (H) that the bullet reaches using the equation:

H = (initial vertical velocity)^2 / (2 * gravity)

H = (565.68 m/s)^2 / (2 * 9.8 m/s^2)
≈ 16341.28 m

So, the maximum height that the bullet reaches is approximately 16341.28 meters.

To find the maximum distance that the bullet could reach, we can use the equation:

D = initial horizontal velocity * time of flight

Since the bullet is fired at an angle of 45 degrees, the initial horizontal velocity (V_x) can be calculated using the equation:

V_x = initial velocity * cos(theta)

V_x = 800 m/s * cos(45 degrees)
≈ 800 m/s * 0.7071
≈ 565.68 m/s

Using these values, we can calculate the maximum distance:

D = 565.68 m/s * 115.32 s
≈ 65153.54 m

So, the maximum distance that the bullet could reach is approximately 65153.54 meters.