A particle is projected from the gr...
Question

# A particle is projected from the ground at an angle with the horizontal. The velocity of the projectile when it is at the greatest height is ( sqrt{frac{3}{5}} ) times when it is at one third of its greatest height. The angle of projection is ( 2 mathrm{K}^{circ} ). Find the value of ( mathrm{K} ).

JEE/Engineering Exams
Physics
Solution
79
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Initial velocity ( = ) drye of projection ( =theta ) Velocity when pasticle is at greatest hoght ucose
[
begin{aligned}
frac{H}{3}=& frac{u^{2} sin ^{2} theta}{6 g}
v_{y}^{2}=& u_{y}^{2}-g_{y} z_{y}
=& u^{2} sin ^{2} theta-2 g frac{u^{2} sin ^{2} theta}{6 g}
&=frac{2}{3} u^{2} sin ^{2} theta
v_{y}=& sqrt{frac{2}{3}} u sin theta
end{aligned}
]
V at higt ( frac{H}{3}=sqrt{frac{5}{3}} ) (U al hight H)
( = )
[
v^{2}=frac{5}{3} u^{2} cos ^{2} theta
]
( Rightarrow u^{2} cos ^{2} theta+frac{2}{3} u^{2} sin ^{2} theta=frac{5}{3} u^{2} cos ^{2} theta )
( Rightarrow quad frac{2}{3} u^{2} sin ^{2} theta=frac{2}{3} u^{2} cos ^{2} theta )
2)
[
tan ^{2} theta=1
]
( Rightarrow quad theta=45^{circ} )
[
begin{array}{l}
2 K^{circ}=45^{circ}
K=frac{45}{2}=22.5^{circ}
end{array}
]

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