A cannon barrel is elevated at an angle of 45°. It fires a ball with a speed of 350 m/s
What is the horizontal range of the cannon?
V^2/g, at that elevation angle.
[2(350^2)cos45sin45]/g = 12.49 m
To determine the horizontal range of the cannon, we can use the basic principles of projectile motion. The horizontal range is the horizontal distance traveled by the ball before it hits the ground.
First, let's break down the velocity of the ball into horizontal and vertical components. Since the cannon barrel is elevated at an angle of 45 degrees, the initial velocity can be split into two components: the horizontal component (Vx) and the vertical component (Vy).
The horizontal component of the velocity (Vx) remains constant throughout the motion since there is no horizontal acceleration. It can be calculated using the formula:
Vx = V * cos(θ)
where V is the initial velocity of the ball and θ is the angle of elevation.
Substituting the given values:
Vx = 350 m/s * cos(45°)
Vx = 350 m/s * 0.7071
Vx = 247 m/s (rounded to three significant figures)
Next, let's find the time it takes for the ball to hit the ground. Since we know the vertical component of the velocity (Vy) and the acceleration due to gravity (g), we can calculate the time using the formula:
t = 2 * Vy / g
In this case, Vy is equal to the initial velocity times the sine of the angle:
Vy = V * sin(θ)
Substituting the given values:
Vy = 350 m/s * sin(45°)
Vy = 350 m/s * 0.7071
Vy = 247 m/s (rounded to three significant figures)
Now, we can find the time:
t = 2 * 247 m/s / 9.8 m/s²
t = 50.41 s (rounded to two decimal places)
Finally, to find the horizontal range (R), we use the formula:
R = Vx * t
Substituting the values:
R = 247 m/s * 50.41 s
R = 12465 m (rounded to three significant figures)
Therefore, the horizontal range of the cannon is approximately 12465 meters.