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Q.(a) State law of parallelogram law of vector addition and prove expression of the formula of the resultant.

(b) Two forces, each equal to F, act on the shown figure, determine Resultant. OR
(a) Define Trajectory of a projectile and hence derive equation of Trajectory of the projectile motion, when projected at an angle (θ) with horizontal motion.
(b) If the horizontal range is four times the maximum height attained by a projectile, then determine the angle of projection.
Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2025Subjective· 5mImportance★★★★★
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The parallelogram law gives the resultant of two vectors as the diagonal of the parallelogram they form; for two equal forces FF at 60° to each other, the resultant comes out to exactly F3F\sqrt3.

(a) Parallelogram Law of Vector Addition: If two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a common point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that same point.

Derivation of the resultant's magnitude: Let vectors A⃗\vec A and B⃗\vec B act at a point with angle θ\theta between them. Complete the parallelogram OACB, with A⃗=OA⃗\vec A = \vec{OA}, B⃗=OB⃗\vec B = \vec{OB}, and resultant R⃗=OC⃗\vec R = \vec{OC} along the diagonal. Drop a perpendicular from C to the extension of OA, meeting it at N.

In right triangle ONC: CN=Bsin⁡θCN = B\sin\theta, ON=A+Bcos⁡θON = A + B\cos\theta.

By Pythagoras in triangle ONC:

R2=ON2+CN2=(A+Bcos⁡θ)2+(Bsin⁡θ)2R^2 = ON^2 + CN^2 = (A+B\cos\theta)^2 + (B\sin\theta)^2

R2=A2+2ABcos⁡θ+B2cos⁡2θ+B2sin⁡2θ=A2+B2+2ABcos⁡θR^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta = A^2 + B^2 + 2AB\cos\theta

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