A particle is projected from the top of a tower of height h with speed u at an angle θ above the horizontal, where 0

Question

A particle is projected from the top of a tower of height h with speed u at an angle θ above the horizontal, where 0<θ<45°.

What is the minimum tower height $h_min$ such that, before or exactly when the projectile reaches the ground, its velocity becomes perpendicular to its initial velocity?

Options

G1 Physics · Accepted Answer

The velocity becomes perpendicular to the initial velocity when u⃗·v⃗ = 0. With u⃗ = u cosθ î + u sinθ ĵ and v⃗ = u cosθ î + (u sinθ-gt)ĵ, we obtain t = u/(g sinθ). At this instant the vertical displacement from the launch point is y = u sinθ·t - ½gt² = -u²cos2θ/(2g sin²θ). Because θ<45°, this displacement is negative. Therefore the projectile must have at least this much vertical space below the launch point. Thus, h m i n h_min h m ​ in = |y| = u²cos2θ/(2g sin²θ).