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Standard VII

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Battles

On an incline...

Question

On an inclined plane two particles $A$ and $B$ are projected with same speed at the same angle with the horizontal, particle $A$ down and particle $B$ up the plane. If the ratio of time of flight of $A$ and $B$ is $cot θ$ , where $θ$ is the angle at which $B$ is projected measured from inclined plane, find the angle (in degree) at which particles are projected

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Solution

From figure,
Time of flight of $A$ ,
$t_{A} = \frac{2 u sin (α + β)}{g cos β}$
Time of flight of $B$ ,
$t_{B} = \frac{2 u sin (α - β)}{g cos β}$
$\frac{t_{A}}{t_{B}} = \frac{sin (α + β)}{sin (α - β)} = \frac{cos θ}{sin θ} = \frac{sin (90^{\circ} - θ)}{sin θ}$
$α + β = 90^{\circ} - θ$
$α - β = θ$
$\Rightarrow α = 45^{\circ}$
Final Answer: 45°

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From figure, Time of flight of A, tA=2usin(α+β)gcosβ Time of flight of B, tB=2usin(α−β)gcosβ tAtB=sin(α+β)sin(α−β)=cosθsinθ=sin(90∘−θ)sinθ α+β=90∘−θ α−β=θ ⇒α=45∘ Final Answer: 45°