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NEET PHYSICSEasy

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is:

A

\theta = tan⁻¹(1/4)

B

\theta = tan⁻¹(4)

C

\theta = tan⁻¹(2)

D

\theta = 45°

Step-by-Step Solution

  1. Formulas:
  • Maximum Height (HmH_m): Hm=v02sin⁡2θ2gH_m = \frac{v_0^2 \sin^2 \theta}{2g} .
  • Horizontal Range (RR): R=v02sin⁡2θgR = \frac{v_0^2 \sin 2\theta}{g} .
  1. Condition: The problem states that the horizontal range equals the maximum height (R=HmR = H_m).
  2. Calculation: v02sin⁡2θg=v02sin⁡2θ2g\frac{v_0^2 \sin 2\theta}{g} = \frac{v_0^2 \sin^2 \theta}{2g} Using the identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta: 2sin⁡θcos⁡θ=sin⁡2θ22 \sin \theta \cos \theta = \frac{\sin^2 \theta}{2} Canceling terms (v02,g,sin⁡θv_0^2, g, \sin \theta) assuming θ≠0\theta \neq 0: 2cos⁡θ=sin⁡θ22 \cos \theta = \frac{\sin \theta}{2} 4=sin⁡θcos⁡θ=tan⁡θ4 = \frac{\sin \theta}{\cos \theta} = \tan \theta θ=tan⁡−1(4)\theta = \tan^{-1}(4)
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