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Find the range of a projectile launched at an angle of 30⁰with an initial velocity of 20m/s

User Jotapdiez
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1 Answer

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Initial velocity of projection (u) = 20 m/s

Angle of projection (θ) = 30°

Formula of Horizontal Range of Projectile:


\boxed{ \bf{R = (2u^2 sin\theta .cos \theta)/(g)}}

By substituting values in the formula, we get:


\rm \longrightarrow R = (2 * 20^2 * sin30 \degree * cos 30 \degree)/(10) \\ \\ \rm \longrightarrow R = \frac{ \cancel{2} * 40 \cancel{0} * ( √(3) )/( 2) * \frac{1}{\cancel{2}} }{ \cancel{10}} \\ \\ \rm \longrightarrow R =40 * ( √(3) )/( 2) \\ \\ \rm \longrightarrow R =20 √(3) \: m \\ \\ \rm \longrightarrow R =34.641 \: m


\therefore Horizontal Range of Projectile (R) = 34.641 m

User Yuby
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