Kinematics

JEE Main level, single correct, medium

159 of 308

Question 159 of 308

A ball is released from a height hh. If t1{t}_{1} and t2{t}_{2} be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t1{t}_{1} and t2{t}_{2}.

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Answer: (D) t2=(2−1)t1{t}_{2}=\left(\sqrt{2}-1\right){t}_{1}

As the body is dropped from a certain height hh, its initial velocity will be zero.
For first half, h1=h2{h}_{1}=\dfrac{h}{2}.
Using second equation of motion, s=ut+12at2s=ut+\dfrac{1}{2}a{t}^{2}
We have, h2=12gt12\dfrac{h}{2}=\dfrac{1}{2}g{t}_{1}^{2}.
Time taken in travel first half of the height, t1=hg...(1){t}_{1}=\sqrt{\dfrac{h}{g}}...\left(1\right).
For total height hh, using again second equation of motion,
h=12g(t1+t2)2⇒(t1+t2)=2hg...(2)h=\dfrac{1}{2}g{\left({t}_{1}+{t}_{2}\right)}^{2}\Rightarrow \left({t}_{1}+{t}_{2}\right)=\sqrt{\dfrac{2h}{g}}...\left(2\right)
From equation (1)\left(1\right) and (2)\left(2\right),
⇒12=t1t1+t2\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{{t}_{1}}{{t}_{1}+{t}_{2}}
⇒1+t2t1=2\Rightarrow 1+\dfrac{{t}_{2}}{{t}_{1}}=\sqrt{2}
⇒t1t2=12−1\Rightarrow \dfrac{{t}_{1}}{{t}_{2}}=\dfrac{1}{\sqrt{2}-1}
Thus, the relation is t2=(2−1)t1{t}_{2}=\left(\sqrt{2}-1\right){t}_{1}.

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