91Ó°ÊÓ

In projectile motion, the horizontal and vertical components of the velocity can be determined using these formulas: \(v_x = v_0 \cos \theta\) \(v_y = v_0 \sin \theta - g t\) where \(v_x\) and \(v_y\) are the horizontal and vertical components of the velocity, \(v_0\) is the initial velocity, \(\theta\) is the angle of projection, \(g\) is the acceleration due to gravity, and \(t\) is the time.

At the greatest height, the vertical component of the velocity is zero, so the overall velocity is \(v_x = v_0 \cos \theta\). At half of the greatest height, the time is not half of the total time of flight, but we can express the vertical component of the velocity in terms of the angle and half of the maximum height (H/2) using the equation: \(v_y = \sqrt{2 g \frac{H}{2}}\) The overall velocity = \(\sqrt{(v_0 \cos \theta)^2 + (\sqrt{2 g \frac{H}{2}})^2}\)

The relation given is: \(v_\text{greatest height} = \sqrt{\frac{2}{5}} v_\text{half-height}\) Substitute the velocities we found in Step 2: \(v_0 \cos \theta = \sqrt{\frac{2}{5}} \sqrt{(v_0 \cos \theta)^2 + (\sqrt{2 g \frac{H}{2}})^2}\)

Square both sides and simplify: \((v_0 \cos \theta)^2 = \frac{2}{5} \left((v_0 \cos \theta)^2 + (2 g \frac{H}{2})\right)\) Divide both sides by \((v_0 \cos \theta)^2\) and solve for \(\cos \theta\): \(\cos \theta = \frac{2 g \frac{H}{2} }{3 (v_0 \cos \theta)^2}\) Now, we can express \(\sin \theta\) using the identity \(\sin \theta = \sqrt{1 - \cos^2 \theta}\), and then divide both sides by \(\cos \theta\) to get \(\tan \theta\): \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{1 - \frac{4 g^2 H^2}{9 (v_0 \cos \theta)^4}}}{\frac{2 g H}{3 (v_0 \cos \theta)^2}}\) Simplify the expression by cancelling out the terms g, H, and \((v_0 \cos \theta)^2\): \(\tan \theta = \frac{\sqrt{1 - \frac{4 g^2 H^2}{9 (v_0 \cos \theta)^4}}} {\frac{2 g H}{3 (v_0 \cos \theta)^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}}\)

Now, we can find the angle of projection by taking the inverse tangent of the expression obtained in Step 4: \(\theta = \tan^{-1} \sqrt{\frac{5}{9}}\) Comparing the value of \(\theta\) with the given options, we find that it matches with Option (C): \(\theta = \tan ^{-1} \frac{2}{3}\) So the correct answer is (C) \(\tan ^{-1} \frac{2}{3}\).