Question

A particle is projected with a velocity v=3i-j+2k and a constant acceleration

Answer 1

Answer:

A particle is projected with a velocity v= (3i – j + 2k) m/s and a constant acceleration acting on the particle is a =( –6i+2j–4k) m/s^2. Then the path of projectile is:- (1) Straight line, (2) Circle, (3) Parabola, (4) Ellipse.11-Jan-2020