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A particle projected from ground with speed u at an angle "a" with the horizontal. Radius of curvature of trajectory of the particle
Dear student, Equation of Trajectory (Path of projectile) At any instant t x= ucosØ.t » t= x/(ucosØ) Also , y= usinØ.t - (1/2)gt2 Substituting for t y= usinØ.x/(ucosØ) - (1/2)g[x/(ucosØ)]2 » y= x.tanØ - [(1/2)g.sec2Ø.x2 ]/u2 Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
Dear student,
Equation of Trajectory (Path of projectile)
At any instant t x= ucosØ.t » t= x/(ucosØ)
Also , y= usinØ.t - (1/2)gt2
Substituting for t y= usinØ.x/(ucosØ) - (1/2)g[x/(ucosØ)]2
» y= x.tanØ - [(1/2)g.sec2Ø.x2 ]/u2
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation. All the best. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
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Sagar Singh
B.Tech, IIT Delhi
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