Question

A body is projected from the ground at an angle of \( \tan^{-1}\sqrt{7} \) with the horizontal. At half of the maximum height, the speed of the body is \( n \) times the speed of projection. The value of \( n \) is

• A. 2
  (2) \( \frac{1}{2} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{3}{4} \) \bigskip
• D.

Hint:

At half the maximum height, the vertical velocity decreases due to gravity, but the horizontal velocity remains unchanged. Using kinematic equations and velocity components, the total speed can be determined.

Answer 1

The Correct Option is D

A body is projected at an angle: \[ \theta = \tan^{-1} \sqrt{7} ) \] with the horizontal. The goal is to determine the speed at half the maximum height in terms of the initial speed. --- Step 1: Components of Initial Velocity Let the initial speed be \( u \). The horizontal and vertical components are: \[ u_x = u \cos \theta, \quad u_y = u \sin \theta. \] From the given angle: \[ \cos \theta = \frac{3}{5}, \quad \sin \theta = \frac{4}{5}. \] Thus: \[ u_x = u \times \frac{3}{5}, \quad u_y = u \times \frac{4}{5}. \] --- Step 2: Maximum Height The maximum height is given by: \[ H = \frac{u_y^2}{2g} = \frac{\left(\frac{4u}{5}\right)^2}{2g} = \frac{16u^2}{50g}. \] Thus, at half the maximum height: \[ h = \frac{H}{2} = \frac{8u^2}{50g}. \] --- Step 3: Vertical Velocity at Half Maximum Height Using the equation of motion: \[ v_y^2 = u_y^2 - 2gh. \] Substituting values: \[ v_y^2 = \left(\frac{4u}{5} \right)^2 - 2g \times \frac{8u^2}{50g}. \] \[ = \frac{16u^2}{25} - \frac{16u^2}{50}. \] \[ = \frac{32u^2}{50} - \frac{16u^2}{50} = \frac{16u^2}{50} = \left(\frac{4u}{10}\right)^2. \] \[ v_y = \frac{4u}{10} = \frac{2u}{5}. \] --- Step 4: Total Speed at Half Maximum Height The horizontal velocity remains the same: \[ v_x = u_x = \frac{3u}{5}. \] The total speed: \[ v = \sqrt{v_x^2 + v_y^2}. \] \[ = \sqrt{\left(\frac{3u}{5} \right)^2 + \left(\frac{2u}{5} \right)^2}. \] \[ = \sqrt{\frac{9u^2}{25} + \frac{4u^2}{25}} = \sqrt{\frac{13u^2}{25}} = \frac{\sqrt{13} u}{5}. \] --- Step 5: Compute \( n \) We define \( n \) as: \[ n = \frac{v}{u} = \frac{\frac{\sqrt{13} u}{5}}{u} = \frac{\sqrt{13}}{5}. \] Approximating \( \sqrt{13} \approx 3.605 \): \[ n \approx \frac{3.605}{5} = 0.72 \approx \frac{3}{4}. \] Thus, the correct answer is: \[ \boxed{\frac{3}{4}} \] which matches option (4).