5.04 Projectile motion examples

Question 1
An object is launched into the air a with an initial velocity of 50ms^-1 at and angle of 30^o.
Calculate:

1. The horizontal distance it travels in 3 secs
2. The height it reaches in
   1. 2 seconds
   2. 3 seconds
   3. 4 seconds
3. Its vertical velocity at:
   1. 2 seconds
   2. 3 seconds
   3. 4 seconds

i)     u_x  =  uCos(Θ)     =  50Cos(30)   =  43.30 ms^-1
       x  =  u_xt    =  43.30 x 3    =  129.90 m

ii)    u_y = uSin(Θ)    = 50Sin(30)    = 25 ms^-1

       y = u_yt - ¹/₂gt²
      a) y = 25 x 2    -   ¹/₂ x 9.81 x 2²     = 30.38m
      b) y = 25 x 3    -   ¹/₂ x 9.81 x 3²     = 30.86m
      c) y = 25 x 4    -   ¹/₂ x 9.81 x 4²     = 21.52m

ii)    v_y = u_y - gt
      a) v_y = 25 - 9.81 x 2 = 5.38 ms^-1
      b) v_y = 25 - 9.81 x 3 = - 4.43 ms^-1
      c) v_y = 25 - 9.81 x 4 = - 14.24 ms^-1

The diagram above summarises the results.
The green arrow represents the horizontal distance travelled at 3 seconds
The blue arrows represent the heights at 2,3, & 4 seconds
The red horizontal arrows represent the constant horizontal velocity of the projectile
The vertical red arrows show the changing magnitude and direction of the projectile's vertical velocity at 2,3 & 4 seconds.