91Ó°ÊÓ

#tag_title# Step 2: Write down the conditions for motion at angle α/2#tag_content# As the body's direction of motion makes an angle α/2 from the horizontal after time t, we have: \(V_x = V_{y,t} \tan \frac{\alpha}{2} \) where \(V_{y,t}\) is the vertical velocity of the body at time t. #tag_title# Step 3: Find the expression for the vertical velocity at time t#tag_content# Using the first equation of motion for vertical motion, we have: \(V_{y,t} = V_y - gt \) Substitute V_y from step 1: \(V_{y,t} = 10\sin \alpha - 10t \) #tag_title# Step 4: Substitute V_y,t and V_x in the condition from step 2 and solve for t#tag_content# Now substitute \(V_x\) and \(V_{y,t}\) in the condition from step 2: \(10 \cos \alpha = (10\sin \alpha - 10t) \tan \frac{\alpha}{2} \) Divide both sides by 10: \(\cos \alpha = (\sin \alpha - t) \tan \frac{\alpha}{2} \) Rearrange the equation to solve for t: \(t = \sin \alpha - \cos \alpha \tan \frac{\alpha}{2} \) Since \(\tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\), we can rewrite the expression for t as: \(t = \sin \alpha - \cos \alpha \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \) Using the trigonometric identity \(\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\), we see that the expression simplifies to: \(t = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} - \cos \alpha \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \) Now, factor out \(\sin \frac{\alpha}{2}\) and simplify the expression: \(t = \sin \frac{\alpha}{2} (2\cos \frac{\alpha}{2} - \frac{\cos \alpha}{\cos \frac{\alpha}{2}}) \) Using the trigonometric identity \(\cos \alpha = 2\cos^2 \frac{\alpha}{2} - 1\), we see that the expression simplifies to: \(t = \sin \frac{\alpha}{2} (2\cos \frac{\alpha}{2} - 1) \) So, the correct answer is (C) \(\sin \frac{\alpha}{2}\).