A vertical pole has a black mark at some height. A stone is projected from a fixed point on the ground. When projected at an angle 45° it hits the pole orthogonally 1 m above the mark. When projected with a different speed at an angle tan⁻¹( $\frac{3}{4}$ ), it hits the pole orthogonally 1.5 m below the mark. Find the angle of projection so that it hits the mark orthogonally to the pole.
Options

Question

A vertical pole has a black mark at some height. A stone is projected from a fixed point on the ground. When projected at an angle 45° it hits the pole orthogonally 1 m above the mark. When projected with a different speed at an angle tan⁻¹( $\frac{3}{4}$ ), it hits the pole orthogonally 1.5 m below the mark. Find the angle of projection so that it hits the mark orthogonally to the pole.

Options

G1 Physics · Accepted Answer

If the projectile strikes a vertical pole orthogonally, its velocity at impact is horizontal, so the pole is reached at the highest point. For a projectile, H/R = tanθ/4, where R is its full horizontal range. Let the black mark be at height H and let R be the common full range corresponding to the fixed pole position. The two given shots give (H + 1)/R = 1/4 and (H - 1.5)/R = ( 3 4 \frac{3}{4} 4 3 ​ )/4 = 3/16. Solving gives H = 9 m and R = 40 m. For the required shot, 9/40 = tanθ/4, hence tanθ…