Kinematics

JEE Main level, numerical answer, medium

238 of 308

Question 238 of 308

A ball of mass mm is thrown vertically upward. Another ball of mass 2m2m is thrown an angle θ\theta with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is 1x\dfrac{1}{x}. The value of xx is _____ .

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Answer: 1

When the first ball of mass mm is projected vertically upwards, the second ball of mass 2m2m makes an angle θ\theta with the vertical.
The time of flight is the same for both the balls.
Then, T1=T2{T}_{1}={T}_{2} ⇒2u1g=2u2sin⁡θg\Rightarrow \dfrac{2{u}_{1}}{g}=\dfrac{2{u}_{2}\sin \theta }{g}
⇒u1=u2sin⁡θ...(1)\Rightarrow {u}_{1}={u}_{2}\sin \theta ...\left(1\right)
Maximum height of projectile is given by H=u2sin⁡2θ2gH=\dfrac{{u}^{2}{\sin }^{2}\theta }{2g}
Ratio of the heights is H1H2=u12u22sin⁡2θ...(2)\dfrac{{H}_{1}}{{H}_{2}}=\dfrac{{u}_{1}^{2}}{{u}_{2}^{2}{\sin }^{2}\theta }...\left(2\right)
From equation (1)\left(1\right) and (2)\left(2\right), we have
⇒H1H2=1\Rightarrow \dfrac{{H}_{1}}{{H}_{2}}=1.
Thus, x=1x=1.

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