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 h1 and h2 respectively, then R = 4\sqrt h_1h_2 Reason R : Product of said heights. h_1h_2 = ( u^2\sin ^2\theta \over 2g )\,.\,( u^2\cos ^2\theta \over 2g ) Choose the correct answer :

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is NOT the correct explanation of A.

Motion in A Plane PYQ — JEE Main 2022 · 25 Jun · Shift 2 · Q46 with Solution

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Correct answer: A

Condition for same range

For a projectile thrown with speed $u$ at angle $\theta$ , the range is $R=\frac{u^2\sin 2\theta}{g}.$

If two projectiles are thrown with the same speed and have the same range, then their angles are complementary: $\theta \quad \text{and} \quad (90^\circ-\theta).$

So let the two balls be projected at angles $\theta$ and $90^\circ-\theta$ .

Maximum heights of the two balls

Maximum height of a projectile is $H=\frac{u^2\sin^2\alpha}{2g}.$

Hence for ball $A$ : $h_1=\frac{u^2\sin^2\theta}{2g}$

and for ball $B$ : $h_2=\frac{u^2\sin^2(90^\circ-\theta)}{2g}= rac{u^2\cos^2\theta}{2g}.$

Therefore, $h_1h_2=\left(\frac{u^2\sin^2\theta}{2g}\right)\left(\frac{u^2\cos^2\theta}{2g}\right).$

So Reason R is true.

Now prove the assertion

From above, $\sqrt{h_1h_2}= rac{u^2\sin\theta\cos\theta}{2g}.$

Thus,

=\frac{2u^2\sin\theta\cos\theta}{g}.$$ Using $\sin 2\theta=2\sin\theta\cos\theta$, $$4\sqrt{h_1h_2}=\frac{u^2\sin 2\theta}{g}=R.$$ Hence, $$R=4\sqrt{h_1h_2}.$$ So **Assertion A is true**. --- 4. **Check whether R explains A** Reason R gives the correct product of heights for complementary angles. From that expression, taking square root and multiplying by $4$ directly leads to $$R=4\sqrt{h_1h_2}.$$ Therefore, **R is the correct explanation of A**. --- 5. **Final choice** Both Assertion and Reason are true, and Reason correctly explains Assertion. $$\boxed{\text{Option A}}$$

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View written solutionFree

Correct answer: A

Condition for same range

For a projectile thrown with speed $u$ at angle $\theta$ , the range is $R=\frac{u^2\sin 2\theta}{g}.$

If two projectiles are thrown with the same speed and have the same range, then their angles are complementary: $\theta \quad \text{and} \quad (90^\circ-\theta).$

So let the two balls be projected at angles $\theta$ and $90^\circ-\theta$ .

Maximum heights of the two balls

Maximum height of a projectile is $H=\frac{u^2\sin^2\alpha}{2g}.$

Hence for ball $A$ : $h_1=\frac{u^2\sin^2\theta}{2g}$

and for ball $B$ : $h_2=\frac{u^2\sin^2(90^\circ-\theta)}{2g}= rac{u^2\cos^2\theta}{2g}.$

Therefore, $h_1h_2=\left(\frac{u^2\sin^2\theta}{2g}\right)\left(\frac{u^2\cos^2\theta}{2g}\right).$

So Reason R is true.

Now prove the assertion

From above, $\sqrt{h_1h_2}= rac{u^2\sin\theta\cos\theta}{2g}.$

Thus,

=\frac{2u^2\sin\theta\cos\theta}{g}.$$ Using $\sin 2\theta=2\sin\theta\cos\theta$, $$4\sqrt{h_1h_2}=\frac{u^2\sin 2\theta}{g}=R.$$ Hence, $$R=4\sqrt{h_1h_2}.$$ So **Assertion A is true**. --- 4. **Check whether R explains A** Reason R gives the correct product of heights for complementary angles. From that expression, taking square root and multiplying by $4$ directly leads to $$R=4\sqrt{h_1h_2}.$$ Therefore, **R is the correct explanation of A**. --- 5. **Final choice** Both Assertion and Reason are true, and Reason correctly explains Assertion. $$\boxed{\text{Option A}}$$

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