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

Question

A particle is projected from the top of a tower with speed u at an angle θ above the horizontal, where 0<θ<45°. At some instant its velocity becomes perpendicular to its initial velocity. The vertical displacement y of the particle from the point of projection at that instant is:

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G1 Physics · Accepted Answer

Initial velocity: u⃗ = u cosθ î + u sinθ ĵ. Velocity after time t: v⃗ = u cosθ î + (u sinθ-gt)ĵ. Perpendicular condition: u⃗·v⃗ = 0. Therefore, u²cos²θ + u sinθ(u sinθ-gt) = 0, u² - ug t sinθ = 0, so t = u/(g sinθ). Vertical displacement is y = u sinθ·t - ½gt² = u²/g - u²/(2g sin²θ) = [u²/(2g)] [2 - 1/sin²θ]. Now, 2sin²θ-1 = -cos2θ. Hence, y = -u²cos2θ/(2g sin²θ). For θ<45°, cos2θ>0, so this point lies below the launch level.

A particle is projected from the top of a tower with speed u at an angle θ above the horizontal, where 0&lt;θ&lt;45°. At some instant its velocity becomes perpendicular to its initial velocity. The vertical displacement y of the particle from the point of projection at that instant is:

A particle is projected from the top of a tower with speed u at an angle θ above the horizontal, where 0<θ<45°. At some instant its velocity becomes perpendicular to its initial velocity. The vertical displacement y of the particle from the point of projection at that instant is: