Question-226249

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Question Number 226249 by fantastic2 last updated on 23/Nov/25

Commented by fantastic2 last updated on 23/Nov/25

Answered by mr W last updated on 24/Nov/25

Commented by mr W last updated on 24/Nov/25

such that the ball reaches point C, ((mu^2 )/R)=mg ⇒u^2 =gR u_A ^2 =u^2 +2gR(1−cos α)=gR(3−2 cos α) AB=2R sin α=((u_A ^2 sin 2α)/g) 2 sin α=(3−2 cos α)sin 2α 1=(3−2 cos α)cos α 2 cos^2 α−3 cos α+1=0 (2 cos α−1)(cos α−1)=0 ⇒cos α=1 ⇒α=0° ⇒rejected ⇒cos α=(1/2) ⇒α=60° ✓

$${such}\:{that}\:{the}\:{ball}\:{reaches}\:{point}\:{C}, \\ $$$$\frac{{mu}^{\mathrm{2}} }{{R}}={mg} \\ $$$$\Rightarrow{u}^{\mathrm{2}} ={gR} \\ $$$${u}_{{A}} ^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{gR}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)={gR}\left(\mathrm{3}−\mathrm{2}\:\mathrm{cos}\:\alpha\right) \\ $$$${AB}=\mathrm{2}{R}\:\mathrm{sin}\:\alpha=\frac{{u}_{{A}} ^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\alpha}{{g}} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\alpha=\left(\mathrm{3}−\mathrm{2}\:\mathrm{cos}\:\alpha\right)\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\mathrm{1}=\left(\mathrm{3}−\mathrm{2}\:\mathrm{cos}\:\alpha\right)\mathrm{cos}\:\alpha \\ $$$$\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{3}\:\mathrm{cos}\:\alpha+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}\:\mathrm{cos}\:\alpha−\mathrm{1}\right)\left(\mathrm{cos}\:\alpha−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\mathrm{1}\:\Rightarrow\alpha=\mathrm{0}°\:\Rightarrow{rejected} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\alpha=\mathrm{60}°\:\checkmark \\ $$

Commented by fantastic2 last updated on 24/Nov/25

$${great}\:{sir} \\ $$

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