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 $θ$ to the horizontal, the maximum height attained by it equals $4 R$ . The angle of projection $θ$ is then given by [NEET 2021]

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

θ

to the horizontal, the maximum height attained by it equals

4 R

. The angle of projection

θ

is then given by [NEET 2021]

Accepted Answer

Correct answer: D. (d) Maximum height attained it is given as H_ max= u^2 sin^2 2g …(i) H_ max=4R (given) Also, we know that, speed of the particle in circular path, u= 2 πRT Substituting the values in the Eq. (i), we get 4R= ( 2 πRT )^2 sin^2 2g ⇒ sin ⁡ = ( 2gT^2 ^2R )^1/2 ⇒ = sin^-1 ⁡ ( 2gT^2 ^2R )^1/2

$A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle θ to the horizontal, the maximum height attained by it equals 4 R . The angle of projection θ is then given by [NEET 2021]$

Solution

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