a projectile is fired at 30 degree above the horizontal with an initial speed of 600 m/s. find the direction and magnitude of the velocity after 2 seconds?

Answer 1

Vy = 300-9.8t
Vx = 519.6

So, at t=2,

Vy = 280.4
Vx = 519.6

|v| = √(280.4^2+519.6^2) = 590.4
θ = arctan(280.4/519.6) = 28.4°

Answer 2

post it.