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
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.