The angle of projection, of which the horizontal range and maximum height of projectile are equal is

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45∘

θ=tan−1⁡(0⋅25)

θ=tan−1⁡4

60∘

answer is C.

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Detailed Solution

We know thatR=u2sin⁡2θg and hmax=u2sin2⁡θ2gGiven that R=hmax∴ u2sin⁡2θg=u2sin2⁡θ2g or 2sin⁡θcos⁡θ=sin2⁡θ/2 or tan⁡θ=4 or θ=tan−1⁡(4)

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