![Q.23 A stationary enemy ship is docked in the sea at a distance of 1.0 km from the coastline. A gun located at the 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/s2] (A) 21.4 (B) 42.9 (C) 23.6 (D) 47.1](data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
Q.23 A stationary enemy ship is docked in the sea at a distance of 1.0 km from the coastline. A gun
located at the 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/s2]
(A) 21.4 (B) 42.9 (C) 23.6 (D) 47.1
Problem Setup
The gun must fire at approximately 21.4° above the horizontal to hit the ship 1 km away with a projectile velocity of 120 m/s. This is determined using the projectile range formula for level ground, as both gun and ship are at sea level. Option (A) is correct, while others yield incorrect ranges .
The range R formula for a projectile launched with initial velocity u at angle θ above horizontal is:
R = \frac{u^2 \sin(2θ)}{g}
Here, R = 1000 m, u = 120 m/s, g = 9.8 m/s². Rearrange to solve for the sine term:
\sin(2θ) = \frac{R g}{u^2} = \frac{1000 \times 9.8}{120^2} = \frac{9800}{14400} = 0.6806
Then, 2θ = \sin^{-1}(0.6806) \approx 42.89^\circ, so θ \approx 21.44^\circ (matches option A: 21.4°).
Option Analysis
Two angles always satisfy the range equation since \sin(2θ) = \sin(180^\circ – 2θ), but the problem asks for the angle above horizontal (typically the lower one for direct fire) .
| Option | θ (°) | 2θ (°) | \sin(2θ) | Matches 0.6806? | Reason |
|---|---|---|---|---|---|
| (A) 21.4 | 21.4 | 42.8 | 0.6794 | Yes (closest) | Correct: R \approx 1000 m . |
| (B) 42.9 | 42.9 | 85.8 | 0.9973 | No | Too high: R \approx 1463 m (excessive range) . |
| (C) 23.6 | 23.6 | 47.2 | 0.7337 | No | Slightly high: R \approx 1052 m (overshoots) . |
| (D) 47.1 | 47.1 | 94.2 | 0.9973 | No | Complementary to (B): R \approx 1463 m (same excessive range) . |
Introduction
Mastering Projectile Motion for Targeting Ships. In physics problems like a gun projectile angle 120 m/s 1 km ship, calculating the optimal elevation uses the range formula R = \frac{u^2 \sin(2θ)}{g}. This gun fires projectile to hit ship scenario tests understanding of two-dimensional kinematics on level ground, common in entrance exams like CSIR NET or JEE. With u=120 m/s, R=1.0 km (1000 m), and g=9.8 m/s², the angle above horizontal is precisely 21.4° .
Step-by-Step Solution
Identify components: Horizontal velocity u_x = u \cos θ, vertical u_y = u \sin θ. Time of flight T = \frac{2 u \sin θ}{g}.
Range: R = u_x T = \frac{u^2 \sin(2θ)}{g}.
Compute \sin(2θ) = 0.6806, so θ = \frac{1}{2} \sin^{-1}(0.6806) \approx 21.4^\circ .
Verify: At 21.4°, range exactly matches 1000 m; higher angles like 42.9° or 47.1° give ~1463 m due to near-90° double angle.
Why Not Other Angles?
42.9° and 47.1° are complementary (sum to 90°), both yielding \sin(2θ) \approx 1 (max range ~1463 m).
23.6° overshoots slightly, as \sin(47.2^\circ) > 0.6806 .
This distinguishes direct (lower θ) vs. lobbing trajectories.
Exam Tips
- Always check units (km to m).
- For max range, θ=45° (\sin(90^\circ)=1), but here it’s specific.
- Practice variations: inclined planes use θ = 45° + α/2.
- Ideal for CSIR NET Life Sciences physics sections or engineering entrances .
