Question

a body projected with a velocity 30m/s at an angle of 30° with the vertical.find the maximum height attained by body​

Answer 1

Topic :- Motion in Plane

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$\maltese \: \underline{\textsf{\textbf{AnsWer :}}}\:\maltese$

✏ A body is projected with a velocity (u) 30m/s at an angle of 30° with the vertical.

✏ We need to find the maximum height attained by body.

✏ The maximum height attained by body is given by ;

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{(u_y)^2}{2g}=\dfrac{u^2\sin^2(\theta)}{2g}$

But here the angle 'θ' is in horizontal ;

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\sin^2(\alpha)}{2g}$

Here, the angle 'α' is in vertical.

Here, the angle made by the α is 30°

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\sin^2( {30}^{ \circ} )}{2g}$

Angle 'θ' can also be written as :

$\qquad \qquad \dag \: \underline{ \footnotesize \sf \theta= 90 - \alpha}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\sin^2( {90}^{ \circ} - \alpha)}{2g}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\sin^2( {90}^{ \circ} -{30}^{ \circ} )}{2g}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\cos^2({30}^{ \circ} )}{2g}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{(30)^2\times\big(\frac{ \sqrt{3} }{2} \big)^{2} }{2 \times 10}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{900\times\frac{3 }{4} }{2 0}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{90\times\frac{3 }{4} }{2}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=45\times\frac{3 }{4}$

$\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\frac{135 }{4}$

$\footnotesize\longrightarrow\:\:\underline{\underline{\sf Maximum \:Height_{(attained \: by \: the \: body)}= 33.75 \:m}}$

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Answer 2

$\huge \blue{\Large{\boxed{\tt{ \red{Answer:- }}}}}$

$\small\sf\orange{Given,}$

$\red\rightarrow$ u = 30m/s

$\rm\purple{Angle\:of\:projection,}$

$\red\rightarrow$ $\small\rm{\theta = 90° - 30° = 60°}$

$\rm\purple{Maximum\: height,}$

$\red\rightarrow$ $\rm{\frac{ {u}^{2} \sin \theta}{2g}}$

$\red\rightarrow$ $\rm{\frac{30 ^{2} \sin ^{2}60°}{2}}$

$\red\rightarrow$ $\rm{\frac{900}{20} \times \frac{3}{4}}$

$\boxed{\tt\blue{Maximum\:height}\pink\implies\frac{135}{4}m=33.75m}$

Hope it helps :)