Question

A particle is projected with initial speed of V₀ and angle of θ. Find the horizontal displacement when its velocity is perpendicular to initial velocity. 

(A) [(V₀²) / (gtanθ)] 

(B) [(V₀²) / (gsinθ)] 

(C) [(V₀ sin θ) / g] 

(D) [(V₀²) / (tanθ)]

Answer 1

Correct option: (A) [(V₀²) / (gtanθ)]

Explanation:

initial velocity vector = V₀ cosθ î + V_o sin θj

Velocity after time t = V_o cos θ î + (V_o sin θ – gt)ĵ

given: velocities are perpendiculars hence dot product is zero

∴ V₀² cos² θ + V₀ sin θ(V₀ sin θ – gt) = 0

∴ V₀² cos² θ + V₀² sin² θ – V₀ sinθ gt = 0

∴ V₀² {(1 – sin θ ∙ gt) / V₀} = 0

∴ 1 = {(sinθ gt) / V₀}

t = {V₀ / (g sinθ)}

Horizontal displacement in time t = (V₀ cos θ)t

hence displacement = (V₀ cos θ) [(V₀) / (g sin θ)]

= [(V₀²) / (gtanθ)]