Question

a boy throws a ball inside a long hall which is 9m high.if a person could throw the ball with maximum speed 10root5 m/s , the maximum horizontal range which would have been obtained under this condition could be?options 40m,46m,48m,50m

Answer 1

Dear Student,


$$\begin{gathered} u=10\sqrt{5}m/s \\ formaximumrangetheangleofthrowmustbe45degree.forthatanglecheckthamaximumheight \\ H=\frac{{\left(10\sqrt{5}\right)}^2{\sin }^{2}45}{2g}=\frac{500}{40}=12.5m \\ henceitisnotpossible. \\ H=9mfindtheta \\ 9=\frac{{\left(10\sqrt{5}\right)}^2{\sin }^2\theta }{2g} \\ \sin \theta =\sqrt{\frac{18g}{500}}=0.6,\cos \theta =0.8 \\ range \\ R=\frac{{\left(10\sqrt{5}\right)}^2\left(2\sin \theta \cos \theta \right)}{g}=\frac{500*2*0.6*0.8}{10}=48m \\ Regards \end{gathered}$$