Hard
Question
A projectile is fired at an angle θ with the horizontal. Find the condition under which it lands perpendicular on an inclined plane inclination α as shown in figure.


Options
A.sin α = cos (θ − α)
B.cos α = sin (θ − α)
C.tan α = cot (θ − α)
D.cot (θ − α) = 2tan α
Solution
Applying equation of motion perpendicular to the incline for y = 0.
0 = V sin (θ − α)t +
(− g cos α) t2

⇒ t = 0 &
At the moment of striking the plane, as velocity is perpendicular to the inclined plane hence component of velocity along incline must be zero.
0 = v cos (θ − α) + ( − g sin α).
v cos (θ − α) = tan α. 2V sin (θ − α)
cot (θ − α) = 2 tan α Ans. (D)
0 = V sin (θ − α)t +

⇒ t = 0 &
At the moment of striking the plane, as velocity is perpendicular to the inclined plane hence component of velocity along incline must be zero.
0 = v cos (θ − α) + ( − g sin α).
v cos (θ − α) = tan α. 2V sin (θ − α)
cot (θ − α) = 2 tan α Ans. (D)
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