Question

A projectile is launched with speed of 128 m/s, at an angle of 60° with the horizontal. After 2.0 s, what is the vertical component of the projectile's velocity?
After 2.0s, what is the speed of the projectile?​

Answer 1

Answer:

a) 91 m/s

b) 111 m/s

Explanation:

v = u + at

v = 128sin60 + (-9.8)(2.0) = 91.25125... m/s

v = √(vx² + vy²) = √((128cos60)² + 91.25125²) = 111.4575... m/s