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30.
How far from the base of the cliff does the cannon ball shot from the top of the cliff travel before it hits the ground.  The cannon ball leaves the cannon 100.0 m above the ground.  It has a velocity of 150. m/s at an inclination of 15.0o.

A.     296 m

B.     306 m

C.     1450 m

D.    1496 m

Down and to the right are positive

Break the inital velocity of the cannon ball into up and forward components
     

vforward =  150. m/s x Cos 15o = 145 m/s

vup = 150. m/s x Sin 15o = 38.8 m/s

   
The vertical displace, velocity, and acceleration determine the time the cannon ball is in the air.

100.0 m = ( - 38.8 m/s ) t   + (1/2)( 9.8 m/s2) t2

4.90 t2 - 38.8 t  -  100.0  =  0            (Use the quadratic equation to solve for t)

t = 9.97 s   and  - 2.05 s

Since the cannon ball cannot hit the ground before it is shot, only the 9.97 s time is valid.

  
Now we use the time with the horizontal velocity to calculate the distance from the base of the cliff.  Remember, there is no acceleration in the horizontal direction for a cannon ball.

X = vt = 145 m/s  x  9.97 s  =  1450 m