Question

A particle is projected from the ground whose \( x, y \)-coordinates vary with time according to the equations \( x = 24t \) and \( y = 43.6t - 4.9t^2 \). Find the initial angle \( \theta \) made by the velocity vector \( \vec{v} \) with the \( x \)-axis.

• A. \( \cot^{-1}(1.82) \)
• B. \( \tan^{-1}(1.82) \)
• C. \( \tan^{-1}(2.82) \)
• D. \( \tan^{-1}(3.4) \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The velocity components in the \( x \) and \( y \) directions are obtained by differentiating the position coordinates with respect to time. The initial angle of projection is the angle of the velocity vector at \( t = 0 \).

Step 2: Key Formula or Approach:
\[ v_x = \frac{dx}{dt}, \quad v_y = \frac{dy}{dt} \] Initial angle \( \theta = \tan^{-1} \left( \frac{v_y(0)}{v_x(0)} \right) \).

Step 3: Detailed Explanation:
1. Differentiate \( x = 24t \): \[ v_x = \frac{d}{dt}(24t) = 24 \] 2. Differentiate \( y = 43.6t - 4.9t^2 \): \[ v_y = \frac{d}{dt}(43.6t - 4.9t^2) = 43.6 - 9.8t \] 3. Find components at \( t = 0 \): \[ v_x(0) = 24 \] \[ v_y(0) = 43.6 \] 4. Calculate the angle: \[ \tan \theta = \frac{43.6}{24} \approx 1.816 \approx 1.82 \] \[ \theta = \tan^{-1}(1.82) \]

Step 4: Final Answer:
The initial angle \( \theta \) is \( \tan^{-1}(1.82) \).