Question

a ball is thrown with speed pf 30 in direction 30* above the horizon what is the height to which it rises
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Answer 1

$\large{\underline{\underline{\red{\sf{\hookrightarrow Given:- }}}}}$

$\sf{A \:body \:is\: thrown\: with \:a\: velocity \:of 30m/s.}$

$\sf It \:is \:projected \:at \:an \:angle\: of\: 30^{\circ}$ .

$\large{\underline{\underline{\red{\sf{\hookrightarrow To\:Find:- }}}}}$

$\sf The \:maximum \:height\: achieved \:by \:body.$

$\large{\underline{\underline{\red{\sf{\hookrightarrow Formula \:Used:- }}}}}$

$\sf When \:a \:body \:when \:thrown\: vertically$

$\sf upward \:with \:some \:angle \:and \:velocity$

$\sf it\: possess\: a \:projectile\: motion\: , then\:$$\sf maximum\: height \:reached\: is\: given \:by\: ,$

$\large\purple{\underline{\boxed{\pink{\sf{\dag Max^m\:height=\dfrac{u^2\:\sin^2\theta}{2g}}}}}}$

$\sf where \:\bf{u }\:is \:velocity\: by\: which\: it\: is \:projected.$

$\large{\underline{\underline{\red{\sf{ \hookrightarrow Calculation:-}}}}}$

$\sf Here ,$

$\green{\bigstar}\orange{\tt Velocity\:of\: projection\:=\:30m/s }$

$\orange{\bigstar}\green{\tt Angle\:of\: projection\:=\:30^{\circ} }$

$\tt\mapsto Now\:put\:the\:values\:in\:above\:stated\: formula:-$

$\tt :\implies H_{max}=\dfrac{u^2\:sin^2\theta}{2g}$

$\tt :\implies H_{max}= \dfrac{(30ms^{-1})^2\times\sin^230^{\circ}}{2\times10ms^{-2}}$

$\tt :\implies H_{max}=\dfrac{30ms^{-1}\times30ms^{-1}\times\bigg(\dfrac{1}{2}\bigg)^2}{20ms^{-2}}$

$\tt :\implies H_{max}=\dfrac{\cancel{30ms^{-1}}\times\cancel{30ms^{-1}}\times1}{\cancel{4}\times\cancel{20ms^{-2}}}$

$\tt:\implies H_{max}=\dfrac{22.5m}{2}$

$\underline{\boxed{\red{\tt{\longmapsto H_{max}\:\:\:\:=\:\:\:\:11.25m}}}}$

$\green{\boxed{\pink{\bf{\leadsto Hence\:the\: maximum\: Height\: attained\:is\:11.25m.}}}}$