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( 3 4 \frac{3}{4} 4 3 ), it hits the pole orthogonally 1.5 m below the mark. Find the required speed of projection (m/s) so that it hits the mark orthogonally to the pole. Take g = 10 m/s².
Physics · Kinematics part-2 Projectile Motion · Projectile from height
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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( 3 4 \frac{3}{4} 4 3 ), it hits the pole orthogonally 1.5 m below the mark. Find the required speed of projection (m/s) so that it hits the mark orthogonally to the pole. Take g = 10 m/s².

Physics · Kinematics part-2 Projectile Motion · Projectile from height

G1 Physics · Accepted Answer

From the two given orthogonal hits, (H + 1)/R = 1/4 and (H - 1.5)/R = 3/16, giving H = 9 m and R = 40 m. For the required shot, tanθ = 4H/R = 9/10, so sinθ = 9/√181 and cosθ = 10/√181. Using R = u²sin2θ/g with g = 10 m/s²: 40 = u²[2(9)(10)/181]/10. Thus u² = 3620/9 and u = √3620/3 ≈ 20.06 m/s.