A cannon barrel is elevated at an angle of 45°. It fires a ball with a speed of 350 m/s. (For the following questions, ignore air resistance.)

(a) What height does the ball reach?
(b) How long is the ball in the air?
(c) What is the horizontal range of the cannon?

a) Solve for V:

(sin45*V)^2^2/2 = (1/4)*V^2 = g H

b ) 2*V sin 45/g

c) 2 sin 45 cos 45 V^2/9 = V^2/g

To solve these questions, we can use the kinematic equations for projectile motion. Let's break it down step by step:

Step 1: Determine the initial velocity components
The initial velocity (V0) of the ball can be broken down into vertical (Vy) and horizontal (Vx) components. Since the cannon is elevated at an angle of 45°, the vertical and horizontal velocities are equal.

Vy = V0 * sinθ
= 350 * sin45°
= 350 * 0.7071
≈ 247.5 m/s

Vx = V0 * cosθ
= 350 * cos45°
= 350 * 0.7071
≈ 247.5 m/s

Step 2: Calculate the time of flight
The time the ball is in the air (T) can be determined using the vertical motion equation:

T = (2 * Vy) / g

Where g is the acceleration due to gravity (approximately 9.8 m/s²).

T = (2 * 247.5) / 9.8
≈ 50.51 seconds

Step 3: Find the maximum height
The maximum height (H) can be found using the formula:

H = (Vy²) / (2 * g)

H = (247.5²) / (2 * 9.8)
≈ 3,096.75 meters

So, the ball reaches a height of approximately 3,096.75 meters.

Step 4: Calculate the horizontal range
The horizontal range (R) can be calculated using the horizontal motion equation:

R = Vx * T

R = 247.5 * 50.51
≈ 12,491.5 meters

Therefore, the horizontal range of the cannon is approximately 12,491.5 meters.

To solve these questions, we can use the principles of projectile motion. Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path due to both horizontal and vertical components of motion.

(a) To find the height the ball reaches, we need to determine the vertical displacement. Given that the cannon is elevated at an angle of 45 degrees, we know that the initial velocity can be decomposed into horizontal (V₀x) and vertical (V₀y) components.

The initial velocity in the vertical direction, V₀y, can be calculated using trigonometry:
V₀y = V₀ * sin(θ),
where V₀ is the initial velocity of the ball and θ is the launch angle.

Substituting the given values, we have:
V₀y = 350 m/s * sin(45°).

The maximum height the ball reaches occurs when its vertical velocity component becomes zero at the top of the trajectory. At this point, the ball starts falling down. To determine the maximum height, we use the equation:

h = (V₀y²) / (2 * g),
where h is the maximum height reached and g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values, we have:
h = (350 m/s * sin(45°))² / (2 * 9.8 m/s²).

Calculating this expression will give us the height the ball reaches.

(b) To find how long the ball is in the air, we can use the time it takes for the ball to reach its maximum height and double it. This is because the time taken to reach the peak is equal to the time taken to fall back down to the initial height.

The time taken to reach the maximum height can be calculated using the formula:

t = V₀y / g.

Once we know the time taken to reach the maximum height, we multiply it by 2 to find the total time in the air.

(c) To calculate the horizontal range, we need to determine the horizontal displacement of the ball. The horizontal displacement is given by:

R = V₀x * t,
where R is the horizontal range, V₀x is the initial horizontal velocity component, and t is the total time in the air.

The initial horizontal velocity component can be obtained using trigonometry:

V₀x = V₀ * cos(θ),
where V₀ is the initial velocity of the ball and θ is the launch angle.

Now that we have all the necessary equations, we can utilize them to find the answers.