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Kinematics

JEE Main level, single correct, medium

246 of 308

Question 246 of 308

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Two identical balls $A$ and $B$ thrown with same velocity ' $u$ ' at two different angles with horizontal attained the same range $R$ . If A and $B$ reached the maximum height ${h}_{1}$ and ${h}_{2}$ respectively, then $R=4\sqrt{{h}_{1}{h}_{2}}$
Reason R: Product of said heights. ${h}_{1}{h}_{2}=\left(\dfrac{{u}^{2}{\sin }^{2}\theta }{2g}\right)\cdot \left(\dfrac{{u}^{2}{\cos }^{2}\theta }{2g}\right)$

Show answer and solution

Answer: (A) Both A and R are true and R is the correct explanation of A.

Range of two balls are

{R}_{1}=\dfrac{{u}^{2}\sin 2{\theta }_{1}}{g}={R}_{2}=\dfrac{{u}^{2}\sin 2{\theta }_{2}}{g}

.
Now, maximum height by both balls are

{h}_{1}=\dfrac{{u}^{2}{\sin }^{2}{\theta }_{1}}{2g}

and

{h}_{2}=\dfrac{{u}^{2}{\cos }^{2}{\theta }_{2}}{2g}

∴\sqrt{{h}_{1}{h}_{2}}=\left(\dfrac{{u}^{2}{\sin }^{2}\theta }{2g}\right)\left(\dfrac{{u}^{2}{\cos }^{2}\theta }{2g}\right)=\dfrac{{u}^{2}\sin \theta \cos \theta }{2g}

=\dfrac{R}{4}

{\Rightarrow R=4\sqrt{{h}_{1}{h}_{2}}}^{}

Hence, both A and R are true and R is the correct explanation of A.

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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