A projectile is launched from the base of a cone and grazes the cone's vertex. The cone has height h and semi-vertical angle θ. Find [u²/(gh) − ½tan²θ]. Use the cone figure.
Physics · Mixed Chapter
JEE Main Practice

Question

A projectile is launched from the base of a cone and grazes the cone's vertex. The cone has height h and semi-vertical angle θ. Find [u²/(gh) − ½tan²θ]. Use the cone figure.

G1 Physics · Accepted Answer

Step 1: The projectile travels horizontally from one edge of the base to the other, so range R = 2h tanθ. Step 2: At the centreline (the cone apex), the trajectory is at its greatest height h. Consequently H_max = u² sin²α/(2g) = h. Step 3: The horizontal range is R = u² sin2α/g = 4h cotα, which implies cotα = ½tanθ. Step 4: u²/(gh) = 2/sin²α = 2(1+cot²α) = 2 + ½tan²θ. Step 5: The requested difference is 2.