Question

A projectile is thrown at an angle of \(60^\circ\) with the horizontal. Initial speed is \(270\, \text{m/s}\). A linear drag force \(F = -CV\) acts on the body. Find the horizontal displacement till \(t = 2\) seconds. Given \(C = 0.1\, \text{s}^{-1}\).

• A. \(135\, \text{m}\)
• B. \(243\, \text{m}\)
• C. \(270\, \text{m}\)
• D. \(260\, \text{m}\)

Hint:

For linear air resistance, use the exponential decay model: \(v = v_0 e^{-Ct}\). Integrate to get displacement.

Answer 1

The Correct Option is B

Step 1: Resolve initial velocity \[ u = 270\, \text{m/s}, \quad \theta = 60^\circ \Rightarrow u_x = u \cos \theta = 270 \cdot \frac{1}{2} = 135\, \text{m/s} \] Step 2: Equation of motion under linear drag force The drag force is proportional to velocity: \[ F = -CV \Rightarrow a = \frac{dv}{dt} = -C v \Rightarrow \frac{dv}{v} = -C dt \] Integrate: \[ \int_{u_x}^{v_x} \frac{dv}{v} = -C \int_{0}^{t} dt \Rightarrow \ln \left( \frac{v_x}{u_x} \right) = -Ct \Rightarrow v_x = u_x e^{-Ct} \] To find horizontal displacement: \[ x(t) = \int_0^t v_x dt = \int_0^t u_x e^{-Ct} dt = \left[ \frac{u_x}{-C} e^{-Ct} \right]_0^t = \frac{u_x}{C} \left(1 - e^{-Ct} \right) \] Substitute values: \[ u_x = 135, \quad C = 0.1, \quad t = 2 \Rightarrow x = \frac{135}{0.1} \left(1 - e^{-0.2} \right) = 1350 \left(1 - e^{-0.2} \right) \] Use: \(e^{-0.2} \approx 0.8187\) \[ x \approx 1350 (1 - 0.8187) = 1350 \times 0.1813 \approx 244 \, \text{m} \] Rounded value: \(\boxed{243\, \text{m}}\)