Projectile Motion

The first task is to break down the original vector of 22.0 m/s with 38.0 degrees into its horizontal and vertical components.

Initial Horizontal Velocity (Vox) =  V(cos A)                Initial Vertical Velocity (Voy) = V(sin A)

= 22.0 Cos 38                                                        = 22.0 Sin 38
= 17.34 m/s                                                            = 13.54 m/s

a. To find hangtime:                                b. To find range:                               c. To find height:
Vfy = Voy + at                                                  d =(Vox)t                                             d=(Voy)t + .5at^2
0  = 13.54 + (-9.8) t                                           = 17.34 (1.3816)                            = 13.54(1.3816) + .5(-9.8)(1.3816)^2
t  = 1.3816 s                                                          = 23.956 m                                         = 9.3537 m
2t = 2.7632 s                                                        = 24.0 m

Projectiles:  Canon

A canon fires an 18.0 kg canon ball with an initial velocity of 22.0 m/s with a angle of 38.0 degrees with ground.  Find a) its hang time, b) its range and c) its maximum height above the ground.  