Final answer:
The horizontal component of the shell's velocity is approximately 1732 m/s, and the vertical component is 1000 m/s. Without air resistance and assuming flat terrain, calculations for range cannot be completed with the given information, and the maximum height would be approximately 51020 m.
Step-by-step explanation:
The question is about a shell being fired at an angle of elevation with a given velocity, and we are asked to find components of the velocity as well as the range and maximum height of the shell. To solve the problem, we will use equations and concepts from projectile motion. Given the velocity (v) is 2 x 103 m/s at a 30-degree angle, the horizontal (vx) and vertical (vy) components of the velocity can be calculated using trigonometric functions: vx = v*cos(θ) and vy = v*sin(θ).
The horizontal component of the velocity is vx = 2 x 103 * cos(30) ≈ 1732 m/s. The vertical component of the velocity is vy = 2 x 103 * sin(30) = 1000 m/s.
To calculate the range (R), we would use the formula R = (v2 * sin(2θ))/g, where g is the acceleration due to gravity (9.8 m/s2). However, this formula is only accurate for flat terrain and neglecting air resistance. The question doesn't provide enough information to calculate the range accurately, as we don't know the height from which the shell is fired or any effects of air resistance.
The maximum height (H) the shell will rise to can be found using the formula H = (vy2)/(2g). This is also an idealized calculation, ignoring air resistance. H = (1000 m/s)2 / (2 * 9.8 m/s2) ≈ 51020 m, under the presumption of no air resistance and that the shell was fired from ground level.