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If R is the range of a projectiel on a horizontal plane and h its maximum height, the maximum horizontal range with the same velocity ofprojection is:

Answer» <html><body><p>2h<br/>`(R^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(8h)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/2r-300472" style="font-weight:bold;" target="_blank" title="Click to know more about 2R">2R</a>+(h^(2))/(8R)`<br/>`2h+(R^(2))/(8h)`</p>Solution :`2h+(R^(2))/(8h)=(2u^(2)<a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a>^(2)theta)/(2g)+(u^(4)sin^(2)theta//<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>^(2))/(8xxu^(2)sin ^(2)theta//2g)` <br/> `=(u^(2))/(g) sin^(2)theta+(u^(2))/(g) cos^(2)theta` <br/> `=(u^(2))/(g)=R_("max")`</body></html>


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