Question

What are the two angles of projection of a projectile projected with velocity 30m/s, so that the horizontal range is 45m. Take, g = 10m/s

CLASS - XI PHYSICS (Kinematics)

Answer 1

Range = $\frac{ v^{2} sin(2 \alpha )}{g}$

45 = tex] \frac{ 30^{2} sin(2 \alpha )}{10} [/tex]

⇒45 = $\frac{900*sin(2 \alpha )}{10}$

⇒ 45 = 90 sin(2α)
⇒sin(2α) = 1/2
⇒ 2α = 30° or 150°
⇒ α = 15° or 75°

Answer 2

Explanation:

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