Question

A particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3}\, ms ^{-1} .$ The angle between velocity vector after $2\, s$ and initial velocity vector is $\left(g=10\, ms ^{-2}\right)$

• A. $0^{\circ}$
• B. $30^{\circ}$
• C. $60^{\circ}$
• D. $90^{\circ}$

Answer 1

Answer: (D)

Initial velocity, $v _{ i }=2 \cos \theta \hat{ i }+4 \sin \theta \hat{ j }=5 \sqrt{3} \hat{ i }+15 \hat{ j}$
Final velocity vector (after $2\, s )$,
$v _{ f }=u \cos \theta \hat{ i }+(u \sin \theta-g t) \hat{ j }=5 \sqrt{3} \hat{ i }-5 \hat{ j }$
Now, $v _{ i } \cdot v _{ f }=25 \times 3-15 \times 5=0$
$\therefore v _{ i } \perp v _{ f }$