Kinematics

JEE Main level, numerical answer, medium

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Question 229 of 308

Two bodies are projected from ground with same speeds 40ms−140m{s}^{-1} at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of 60∘60^{\circ}, with horizontal then sum of the maximum heights, attained by the two projectiles, is _____ mm. (Given g=10ms−2g=10m{s}^{-2})

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Answer: 80

For two projectiles with same speed to have the same horizontal range, angle of projection will be θ1+θ2=90∘{\theta }_{1}+{\theta }_{2}=90^{\circ}.
Therefore, θ2=90∘−60∘=30∘{\theta }_{2}=90^{\circ}-60^{\circ}=30^{\circ}.
Now, required sum of maximum heights will be,
(Hmax⁡)1+(Hmax⁡)2=u2sin⁡2θ12g+u2sin⁡2θ22g{\left({H}_{\max }\right)}_{1}+{\left({H}_{\max }\right)}_{2}=\dfrac{{u}^{2}{\sin }^{2}{\theta }_{1}}{2g}+\dfrac{{u}^{2}{\sin }^{2}{\theta }_{2}}{2g}
=40220(14+34)=80m=\dfrac{{40}^{2}}{20}\left(\dfrac{1}{4}+\dfrac{3}{4}\right)=80m

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