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What are the two angles of projection of a projectile projected with velocity of \({30}{m}/{s}\), so that the horizontal range is \({45}{m}\). Take, \({g}={10}{m}/{s}^{{{2}}}\).
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What are the two angles of projection of a projectile projected with velocity of \({30}{m}/{s}\), so that the horizontal range is \({45}{m}\). Take, \({g}={10}{m}/{s}^{{{2}}}\).

Answer

Horizontal range \(=\frac{{{u}^{{{2}}}{\sin{{2}}}\theta}}{{g}}\)45=(30^^(2) sin 2 theta)/(10)\(\)sin 2 theta =(45 xx 10)/(30 xx 30) =1/2 =sin 30^(@) \({\quad\text{or}\quad}{{\sin{{15}}}^{{\circ}}}\)
\({2}\theta={30}^{{\circ}}{\quad\text{or}\quad}{15}^{{\circ}}{\quad\text{or}\quad}\theta={15}^{{\circ}}{\quad\text{or}\quad}{75}\%{\left(\circ\right)}\).
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