A particle is projected with certain velocity at two different angles of projections with respect to horizontal plane so as to have same range $R$ on a horizontal plane. If $t_{1}$ and $t_{2}$ are the time taken for the two paths, then which one of the following relations is correct?

t_{1} t_{2} = \frac{2 R}{g}

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t_{1} t_{2} = \frac{R}{g}

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t_{1} t_{2} = \frac{R}{2 g}

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t_{1} t_{2} = \frac{4 R}{g}

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Solution

The correct option is B $t_{1} t_{2} = \frac{2 R}{g}$
Horizontal range is same for angle of projection θ and 90o−θ
∴ Horizontal range R=u2sin2θg...(i)
∴t1=2usinθg,t2=2usin(90o−θ)g=2ucosθg
∴t1t2=4u2sinθcosθg2=2g(2u2sinθcosθg)
=2g(u2sin2θg)(∴sin2θ=2sinθcosθ)
=2Rg (Using (i))

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A particle is projected with certain velocity at two different angles of projections with respect to horizontal plane so as to have same range R on a horizontal plane. If t1 and t2 are the time taken for the two paths, then which one of the following relations is correct?

A particle is projected with certain velocity at two different angles of projections with respect to horizontal plane so as to have same range $R$ on a horizontal plane. If $t_{1}$ and $t_{2}$ are the time taken for the two paths, then which one of the following relations is correct?