# 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)

$\begin{array}{l} R=\frac{u^{2} \sin 2 \theta}{g}=\frac{(30)^{2} \sin 2 \theta}{10}=45 \\ \Rightarrow \sin 2 \theta=\frac{450}{(30)^{2}} \\ \Rightarrow \sin 2 \theta=1 / 2 \\ \Rightarrow 2 \theta=30^{\circ} \text { or } 150^{\circ} \Rightarrow \theta=15^{\circ} \text { or } 75^{\circ} \end{array}$

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