ઉદગમ સ્થાનથી t = 0 સમયે ફેંકેલ પદાર્થ નું સ્થ | vedclass

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Question

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,From\,question,\
\,\,\,\,\,\,\,\,\,Horizontal\,velocity\,\left( {initial} ight),\
\,\,\,\,\,\,\,\,\,{u_x} =

Vedclass · Accepted Answer

${	an ^{ - 1}}\frac{7}{4}$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,From\,question,\
\,\,\,\,\,\,\,\,\,Horizontal\,velocity\,\left( {initial} ight),\
\,\,\,\,\,\,\,\,\,{u_x} = \frac{{40}}{2} = 20m/s\
\,\,\,\,\,\,\,\,\,Vertical\,velocity\,\left( {initial} ight),\
\,\,\,\,\,\,\,\,\,\,50 = {u_y}t + \frac{1}{2}g{t^2}\
 \Rightarrow \,{u_y} 	imes 2 + \frac{1}{2}\left( { - 10} ight) 	imes 4\
or,\,50 = 2{u_y} - 20
\end{array}$

$\begin{array}{l}
or,\,\,{u_y} = \frac{{70}}{2} = 35m/s\
	herefore \,	an \,	heta  = \frac{{{u_y}}}{{{u_x}}} = \frac{{35}}{{20}} = \frac{7}{4}\
 \Rightarrow \,\,Angle\,	heta {	an ^{ - 1}}\frac{7}{4}
\end{array}$

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