Kinematics

JEE Main level, single correct, medium

223 of 308

Question 223 of 308

Two projectiles are projected at 30∘30^{\circ} and 60∘60^{\circ}with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

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Answer: (D) 1:31:3

The maximum height is given by, H=u2sin⁡2θ2gH=\dfrac{{u}^{2}{\sin }^{2}\theta }{2g}.
The given data is
θ1=30∘θ2=60∘u1=u2{\theta }_{1}=30^{\circ} {\theta }_{2}=60^{\circ} {u}_{1}={u}_{2}
Therefore,
(H1)max(H2)max=u12sin⁡2θ1u22sin⁡2θ2=(sin⁡30∘sin⁡60∘)2=(1232)2=13\dfrac{{\left({H}_{1}\right)}_{max}}{{\left({H}_{2}\right)}_{max}}=\dfrac{{u}_{1}^{2}{\sin }^{2}{\theta }_{1}}{{u}_{2}^{2}{\sin }^{2}{\theta }_{2}} ={\left(\dfrac{\sin 30^{\circ}}{\sin 60^{\circ}}\right)}^{2} ={\left(\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}^{2} =\dfrac{1}{3}

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