Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ph-1765: NEET 2012 Physics - Class 12 AIPM-2012-Ph-1765 Solution: The horizontal range and the maximum height of a projectile are …

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is θ=tan⁡−1(14)\theta = \tan^{-1}(\frac{1}{4}) θ=tan⁡−1(4)\theta = \tan^{-1}(4) θ=tan⁡−1(2)\theta = \tan^{-1}(2) θ=45∘\theta = 45^\circ

Multiple Choice Options

  • A 24hR25m{{24hR} \over {25m}} 25 m 24 h R _
  • B 25hR24m{{25hR} \over {24m}} 24 m 25 h R _
  • C 25m24hR{{25m} \over {24hR}} 24 h R 25 m _
  • D 24m25hR{{24m} \over {25hR}} 25 h R 24 m _

Step-by-Step Verified Solution

We know, 1 _ = R Z 2 ( 1 n 1 2 _ 1 n 2 2 ) where R = Rydberg constant, Z = atomic number Here, n 1 = 1, n 2 = 5 _ 1 _ = R ( 1 1 2 _ 1 5 2 ) = 24 R 25 According to conservation of linear momentum, we get Momentum of photon = Momentum of atom _ h _ = mv _ v = h m _ = h m ( 24 R 25 ) = 24 h R 25 m

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