Question

A particle is projected from a point on the ground with an initial velocity of u - 50 m/s at an angle of 530
with the horizontal (tan 53° - 4/3,9 -10 m/s2 = acceleration due to gravity).
(A) The velocity of the particle will make angle 45" with the horizontal after time 1 s
(B) The velocity of the particle will make angle 45° with the horizontal after time 7 s.
(C) The average velocity between the point of projection and the highest point on its path is horizontal
(D) The average velocity between two points on same height will be horizontal.​

Answer 1

Vx=u cos theta

=50 ×( 3/5)

=30m/s

Vy=u sin theta - gt

= 50 × (4/5) - 10×1

=40-10

=30m/s

Tan theta = Vy÷Vx

=30/30

=1

ThetA = 45°

So option A is correct.