Physics Circle home Student Login

Home›Chapters›Projectile Motion PYQs›Time of Flight, Maximum Height & Range›Question 2

MHT CET 2026 (18 Apr, Shift 2) · Time of Flight, Maximum Height & Range

Time of Flight, Maximum Height & Range question from MHT CET 2026 (18 Apr, Shift 2)

Q2MHT CET 2026 · 18 APR · SHIFT 2Easy

Time of Flight, Maximum Height & Range

For a projectile motion, the range R is 'n' times the maximum height H. So the angle of projection is

Show answer and solution

Answer: (c)

  1. RH=2u2sin⁡θcos⁡θ/gu2sin⁡2θ/2g=4tan⁡θ\frac{R}{H} = \frac{2u^2\sin\theta\cos\theta/g}{u^2\sin^2\theta/2g} = \frac{4}{\tan\theta}
  2. 4tan⁡θ=n⇒tan⁡θ=4n\frac{4}{\tan\theta} = n \Rightarrow \tan\theta = \frac{4}{n}
  3. So the answer is (c) tan⁡−1(4n)\tan^{-1}\left(\frac{4}{n}\right).
Report a mistake

More questions on Time of Flight, Maximum Height & Range