A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle '$\theta$' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $\theta$, is then given by :

to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $\\theta$, is then given by :","description":"Practice this DASH question: A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were pr. Tag details: NEET: NEET exam practice question for medical entrance preparation; JEE Advanced: JEE Advanced practice topic for deeper concept preparation; JEE Mains: JEE Main practice topic for engineering entrance preparation; KCET: KCET exam practice question for Karnata.","url":"https://learnwithdash.ai/portal/question/792604/","keywords":"NEET, JEE Advanced, JEE Mains, KCET, Motion in a Plane, Circular Motion, Projectile Motion, NCERT, NEET Physics, Class 11 Physics, Class 12 Physics, Physics, Chemistry, Biology","about":[{"@type":"Thing","name":"NEET","description":"NEET: NEET exam practice question for medical entrance preparation"},{"@type":"Thing","name":"JEE Advanced","description":"JEE Advanced: JEE Advanced practice topic for deeper concept preparation"},{"@type":"Thing","name":"JEE Mains","description":"JEE Mains: JEE Main practice topic for engineering entrance preparation"},{"@type":"Thing","name":"KCET","description":"KCET: KCET exam practice question for Karnataka entrance preparation"},{"@type":"Thing","name":"Motion in a Plane","description":"Motion in a Plane: DASH practice topic connected to this question, quiz, and solution"},{"@type":"Thing","name":"Circular Motion","description":"Circular Motion: DASH practice topic connected to this question, quiz, and solution"},{"@type":"Thing","name":"Projectile Motion","description":"Projectile Motion: DASH practice topic connected to this question, quiz, and solution"},{"@type":"Thing","name":"NCERT","description":"NCERT: NCERT-aligned concept practice and textbook-based preparation"},{"@type":"Thing","name":"NEET Physics","description":"NEET Physics: NEET exam practice question for medical entrance preparation"},{"@type":"Thing","name":"Class 11 Physics","description":"Class 11 Physics: Class 11 syllabus topic for board and competitive exam practice"},{"@type":"Thing","name":"Class 12 Physics","description":"Class 12 Physics: Class 12 syllabus topic for board and competitive exam practice"},{"@type":"Thing","name":"Physics","description":"Physics: Physics concept practice with question and solution review"},{"@type":"Thing","name":"Chemistry","description":"Chemistry: Chemistry concept practice with question and solution review"},{"@type":"Thing","name":"Biology","description":"Biology: Biology concept practice with question and solution review"}]}

Your AI-Powered Personal Tutor

Question

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution.

If this particle were projected with the same speed at an angle '$\theta$' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $\theta$, is then given by :

$\theta = {\sin ^{ - 1}}{\left( {{{2g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}}$

$\theta = {\cos ^{ - 1}}{\left( {{{g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}}$

$\theta = {\cos ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}}$

$\theta = {\sin ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}}$

Solution

Please login to view the detailed solution steps...

Go to DASH