![3. A stationary enemy ship is docked in the sea at a distance of 1.0 km from the coastline. A gun located at sea level on the coastline can fire projectiles at a velocity of 120 m/s. What is the angle (in degrees) above the horizontal at which the gun must fire to hit the ship? [g = 9.8 m s⁻²] (A) 21.4 (B) 42.9 (C) 23.6 (D) 47.1](https://www.letstalkacademy.com/wp-content/uploads/2026/07/1-3-6.jpg)
3. A stationary enemy ship is docked in the sea at a distance of 1.0 km from the coastline. A gun located at sea level on the coastline can fire projectiles at a velocity of 120 m/s. What is the angle (in degrees) above the horizontal at which the gun must fire to hit the ship? [g = 9.8 m s⁻²]
(A) 21.4
(B) 42.9
(C) 23.6
(D) 47.1
Projectile Motion Angle to Hit a Ship 1 km Away: Detailed Solution
Understanding the Projectile Motion Problem
This is a standard projectile motion problem involving the horizontal range of a projectile. The gun and the stationary ship are both located at sea level, which means that the projectile is fired and hits the target at the same vertical level. Therefore, the standard formula for the horizontal range of a projectile can be applied directly.
The ship is located 1.0 km from the coastline. Since all quantities must be expressed in SI units, the distance should first be converted from kilometres to metres. Therefore, the horizontal range is R = 1.0 km = 1000 m. The initial velocity of the projectile is u = 120 m/s, while the acceleration due to gravity is g = 9.8 m/s2.
Formula for the Horizontal Range of a Projectile
When a projectile is launched with an initial velocity u at an angle θ above the horizontal and lands at the same vertical level from which it was projected, its horizontal range is given by:
R = (u2 sin 2θ)/g
In this equation, R represents the horizontal range, u is the initial velocity of projection, θ is the angle of projection above the horizontal, and g is the acceleration due to gravity.
Substituting the Given Values
For the projectile to hit the ship, its horizontal range must be exactly 1000 m. Substituting R = 1000 m, u = 120 m/s and g = 9.8 m/s2 into the range formula gives:
1000 = [(120)2 sin 2θ]/9.8
Since (120)2 = 14400, the equation becomes:
1000 = (14400 sin 2θ)/9.8
Rearranging the equation to calculate sin 2θ:
sin 2θ = (1000 × 9.8)/14400
sin 2θ = 9800/14400
sin 2θ ≈ 0.6806
Calculating the Angle of Projection
Taking the inverse sine on both sides gives:
2θ = sin−1(0.6806)
2θ ≈ 42.9°
Therefore:
θ = 42.9°/2
θ ≈ 21.45°
Rounding the value according to the given options, the required angle of projection is approximately 21.4°.
Why Is Option (A) 21.4° Correct?
Option (A) is correct because substituting θ = 21.4° into the projectile range equation produces a horizontal distance very close to 1000 m. At this angle, 2θ = 42.8°, and sin 42.8° is approximately 0.68. Therefore, the range becomes approximately (1202 × 0.68)/9.8, which is close to 1000 m. Hence, the projectile reaches the stationary ship located 1.0 km away.
Detailed Analysis of All Options
Option (A): 21.4°
This is the correct answer. Using the horizontal range equation, the required value of sin 2θ is approximately 0.6806. The corresponding acute value of 2θ is approximately 42.9°, which gives θ ≈ 21.4°. Therefore, a projectile fired at approximately 21.4° above the horizontal can hit the ship located 1000 m away.
Option (B): 42.9°
This option is incorrect because 42.9° is approximately the value of 2θ, not θ. The range formula contains sin 2θ, so after calculating 2θ = 42.9°, the value must be divided by 2 to obtain the actual angle of projection. Thus, the required angle is approximately 21.4°, not 42.9°.
Option (C): 23.6°
This option is incorrect because substituting θ = 23.6° into the range equation does not give a horizontal range of exactly 1000 m. For this angle, 2θ = 47.2°, and the corresponding value of sin 2θ is greater than the required value of approximately 0.6806. Therefore, the projectile would travel farther than the required distance.
Option (D): 47.1°
This option is also incorrect. If the projectile is fired at 47.1°, then 2θ = 94.2°. The value of sin 94.2° is close to 1, producing a horizontal range close to the maximum possible range for the given initial velocity. Since the ship is only 1000 m away, this angle does not satisfy the required range condition.
Important Concept of Complementary Angles in Projectile Motion
For a given initial velocity, two different angles of projection can produce the same horizontal range, provided the projectile starts and lands at the same level. These two angles are complementary, meaning their sum is 90°.
Since the lower angle obtained here is approximately 21.4°, another possible theoretical angle is approximately:
90° − 21.4° = 68.6°
Thus, both approximately 21.4° and 68.6° can produce the same horizontal range of 1.0 km. However, only 21.4° is present among the given options, so option (A) is the required answer.
Verification Using the Maximum Range
The maximum horizontal range of a projectile is obtained when the angle of projection is 45°. The maximum range is given by:
Rmax = u2/g
Substituting u = 120 m/s and g = 9.8 m/s2:
Rmax = 14400/9.8 ≈ 1469.4 m
Since the ship is located only 1000 m away, which is less than the maximum possible range of approximately 1469.4 m, the projectile can easily reach the target. This calculation also confirms that two possible projection angles exist for this range.
Final Answer
Correct Option: (A) 21.4°
The gun must fire the projectile at an angle of approximately 21.4° above the horizontal to hit the stationary ship located 1.0 km from the coastline. This result is obtained by applying the horizontal range formula R = u2 sin 2θ/g and solving for the angle of projection.
