Final answer:
The initial velocity of the shell is approximately 1100 m/s. The shell does not reach any maximum height. The surface of the ocean will be approximately 63.43 m lower 32.0 km from the ship. Therefore, the correct answer is option d) 1100 m/s
Step-by-step explanation:
To calculate the initial velocity of the shell, we can use the standard equations of projectile motion. The maximum distance of 32.0 km is the horizontal range of the shell. We can use the equation:
R = (v02*sin(2θ))/g
Where R is the range, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. Since the shell is fired horizontally, the launch angle θ can be taken as 0 degrees. Rearranging the equation, we have:
v0 = √(R*g/sin(2θ))
Substituting the given values, we have v0 ≈ 1100 m/s.
To calculate the maximum height reached by the shell, we can use the equation:
H = (v02*sin2(θ))/(2g)
Since the shell is fired horizontally, the launch angle θ is 0 degrees. Substituting the given values, we have H ≈ 0 m.
To calculate how many meters lower the surface of the ocean will be 32.0 km from the ship, we can use the equation:
h = R2 / (2Rearth)
Where h is the height difference, R is the distance from the ship (32.0 km), and Rearth is the radius of the Earth (6.37 × 103 km).
Substituting the values, we have h ≈ 63.43 m.
The error introduced by assuming a flat Earth in projectile motion is not significant here, as the height difference of 63.43 m is relatively small compared to the range of 32.0 km.
Therefore, the correct answer is option d) 1100 m/s