Question

A particle is projected from an origin o ay a velocity of (3i + 13j)m/s. Find the position vector of the particle 2 seconds after projection and the distance the particle is then from o. (Take g=10 m/s^2

Answer 1

Answer:

x position = 6i

y position = 13×2-1/2 ×10×4

=26-20

=6j

so position after 2 seconds = 6i +6j

hence distance from O

=sq root( 36 +36)

=sq root(76)

= 8.7m