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} ).

Solution

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}

]