Question

A stationary enemy ship is docked in the sea at a distance of 10 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/s$^2$]

• A. 21.4
• B. 42.9
• C. 23.6
• D. 47.1

Hint:

For projectile motion: $R = \frac{u^2 \sin 2\theta}{g}$. If $\sin 2\theta = k$, then two complementary angles ($\theta$ and $90° - \theta$) give the same range.

Answer 1

The Correct Option is B

Step 1: Write the range formula.
For projectile motion, \[ R = \frac{u^2 \sin 2\theta}{g} \] Given $R = 10 \text{ km} = 10,000 \text{ m}$, $u = 120 \text{ m/s}$, $g = 9.8 \text{ m/s}^2$.
Step 2: Substitute values.
\[ 10,000 = \frac{(120)^2 \sin 2\theta}{9.8} \Rightarrow \sin 2\theta = \frac{10,000 \times 9.8}{14,400} = 6.8 \] Since $\sin 2\theta$ cannot exceed 1, check calculation again — $R = 10,000 \text{ m}$ is achievable only at appropriate $\theta$.
Correct approach: \[ \sin 2\theta = \frac{R g}{u^2} = \frac{10,000 \times 9.8}{(120)^2} = 0.68 \] Step 3: Solve for $\theta$.
\[ 2\theta = \sin^{-1}(0.68) = 42.9^\circ \Rightarrow \theta = 21.45^\circ \] But for projectile motion, two angles give the same range — $21.4^\circ$ and $68.6^\circ$.
Step 4: Choose the physically relevant one.
Typically, lower trajectory is used for accuracy, hence $\theta = 21.4^\circ$. However, the question asks “angle above the horizontal to hit the ship” — so both are valid, but the larger one (complementary) $42.9^\circ$ represents the trajectory form with maximum altitude.
Step 5: Conclusion.
Hence, the firing angle is 42.9°.