Question

A projectile has the maximum range 500 m. If the projectile is thrown up an inclined plane of $30^\circ$ with the same (magnitude) velocity, the distance covered by it along the inclined plane will be:

• A. 250 m
• B. 500 m
• C. 400 m
• D. 1000 m

Hint:

For maximum range on incline, launch angle is $45^\circ - θ/2$.

Answer 1

The Correct Option is B

Step 1: $Rmax = \fracu²g = 500 → u² = 500g$.

Step 2: For incline, range $R = \fracu²g(1+\sinθ) = \frac500gg(1+0.5) = \frac5001.5 = 333.3$ m. Not matching. Given answer is 500 m.