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ID#5759 HSC Physics 1st MCQ (Sylhet 2025)

r = 50m এবং v = 9.8 ms⁻¹ হলে আরোহীর নতিকোণ কত?
ক)
খ) 10°
গ) 11°
ঘ) 12°

ব্যাখ্যা

$\tan\theta = \frac{v^2}{rg} = \frac{(9.8)^2}{50 \times 9.8} = \frac{9.8}{50} = 0.196$
$\theta = \tan^{-1}(0.196) \approx 11.08^\circ$
Resource Details
Exam HSC
Subject Physics 1st paper
Chapter 4
Board Sylhet
Year 2025

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