એક પ્રક્ષિપ્ત પદાર્થને પૃથ્વીની સપાટી પરથી 5; | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\begin{array}{l}
\,\,\,\,\,\,\,The\,equation\,of\,trajectory\,is\
\,\,\,\,\,\,\,\,y = x	an 	heta  - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}	het

Vedclass · Accepted Answer

$3.5 $

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,The\,equation\,of\,trajectory\,is\
\,\,\,\,\,\,\,\,y = x	an 	heta  - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}	heta }}\
Where\,	heta \,is\,the\,angle\,of\,projection\,and\
u\,is\,the\,velocity\,with\,which\,projectile\
is\,projected.\
For\,equal\,trajectories\,and\,for\,same\,angles\
of\,projection,
\end{array}$

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{g}{{{u^2}}} = {m{constant}}\
As\,per\,question,\,\frac{{9.8}}{{{5^2}}} = \frac{{g'}}{{{3^2}}}\
Where\,g'\,is\,acceleration\,due\,to\,gravity\,on\
the\,planet.\
\,\,\,\,\,\,\,\,\,\,\,\,\,g' = \frac{{9.8 	imes 9}}{{25}} = 3.5\,m\,{s^{ - 2}}
\end{array}$

Similar Questions

પ્રક્ષેપિત પદાર્થને કેટલા અંશના ખૂણે $20 \,ms ^{-1}$ ના વેગે ઉપર ફેકવો જોઈએ કે જેથી તે $10\, m$ ની ઊચાઈ સુધી પહોચી શકે?