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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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## QuestionPhysicsClass 11

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, $${g}={10}{m}/{s}^{{{2}}}$$.

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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