MCQEasyJEE 2023Projectile Motion

JEE Physics 2023 Question with Solution

Given below are two statements: Assertion (A) and Reason (R). Assertion A: When a body is projected at an angle of 4545^\circ, its range is maximum. Reason R: For maximum range, the value of sin2θ\sin 2\theta should be equal to 11.

  • A

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

  • B

    A is false but R is true

  • C

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

  • D

    A is true but R is false

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A projectile is projected with speed uu at angle θ\theta.

Find: Whether Assertion (A) and Reason (R) are true, and whether R correctly explains A.

For the range of a projectile,

R=u2sin(2θ)gR = \frac{u^2 \sin(2\theta)}{g}

For maximum range, the factor sin(2θ)\sin(2\theta) must be maximum.

The maximum value of sin(2θ)\sin(2\theta) is 11. So,

sin(2θ)=1\sin(2\theta) = 1

which gives

2θ=902\theta = 90^\circ

and therefore,

θ=45\theta = 45^\circ

Hence, the range is maximum when the angle of projection is 4545^\circ.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). The correct option is C.

Using the range expression

Given: The statement about maximum range of a projectile and the reason involving sin(2θ)\sin(2\theta).

Find: The correct assertion-reason option.

The assertion is true because the range of a projectile is given by

R=(v2g)sin(2θ)R = \left(\frac{v^2}{g}\right) \sin(2\theta)

To maximize RR, we must maximize sin(2θ)\sin(2\theta). A maximum occurs when

cos(2θ)=0\cos(2\theta) = 0

Thus,

2θ=(2n+1)π22\theta = (2n+1)\frac{\pi}{2}

so

θ=(2n+1)π4\theta = (2n+1)\frac{\pi}{4}

For the physically relevant acute angle giving maximum range,

θ=45\theta = 45^\circ

Thus, the reason is true and it correctly explains why the assertion is true.

Therefore, the correct option is C.

Common mistakes

  • Students often confuse maximum height with maximum range. A projectile has maximum range at 4545^\circ, but maximum height increases for larger projection angles at the same speed. Use the range formula, not the height formula.

  • A common mistake is to write sin(2θ)=90\sin(2\theta)=90^\circ. The sine of an angle is a number, so the correct statement is 2θ=902\theta=90^\circ when sin(2θ)=1\sin(2\theta)=1. First identify the angle, then solve for θ\theta.

  • Some students forget that the maximum value of sinx\sin x is 11. If this is missed, they cannot identify the condition for maximum range. Always inspect the trigonometric factor in the expression and use its maximum possible value.

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