a body projected with a velocity 30m/s at an angle of 30° with the ve

1.	a body projected with a velocity 30m/s at an angle of 30° with the vertical.find the maximum height attained by body
Answer» Topic :- Motion in Plane $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ $\maltese \: \underline{\textsf{\textbf{AnsWer :}}}\:\maltese$ ✏ 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 : $\qquad \qquad \dag \: \underline{ \footnotesize \sf \theta= 90 - \alpha}$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\sin^2( {90}^{ \circ} - \alpha)}{2g} \\$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\sin^2( {90}^{ \circ} -{30}^{ \circ} )}{2g} \\$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{u^2\cos^2({30}^{ \circ} )}{2g} \\$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{(30)^2\times\big(\frac{ \sqrt{3} }{2} \big)^{2} }{2 \times 10} \\$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{900\times\frac{3 }{4} }{2 0} \\$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\dfrac{90\times\frac{3 }{4} }{2} \\$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=45\times\frac{3 }{4} \\$ $\footnotesize\longrightarrow\:\:\sf Maximum \:Height_{(attained \: by \: the \: body)}=\frac{135 }{4} \\$ $\footnotesize\longrightarrow\:\:\underline{\underline{\sf Maximum \:Height_{(attained \: by \: the \: body)}= 33.75 \:m}} \\$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

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

Answer»

Topic :- Motion in Plane

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

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

✏ 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 :

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

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

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

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

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

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

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

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

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

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

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}$