In: Physics
2. You are firing a cannonball from the top of a cliff at a pirate ship which is behind a mountain. The height of the cannon is 100 m, the height of the mountain is 120 m, and the height of the pirate ship is 70 m.
The mountain is horizontally 200 m away from cannon, while the pirate ship is horizontally 270 m away from the cannon.
The speed of the cannonball after it is fired is 65 m s-1. You could hit anywhere from the bottom of the pirate ship to the top in order to sink it. Over what angles could you fire the cannonball to sink the pirate ship?
X-component of speed of cannonball = 65 cos
Y-component of speed of cannonball = 65 Sin
Along the x-direction , P to Q motion :::
horizontal distance = AC = 270 m
let the time taken be ''t''
then , 65 cos t = 270
t = 270 / ( 65 cos) Eq-1
Along the y-direction , P to Q motion :::
vertical distance = d = 70 - 100 = -30 m
acceleration = a = g = -9.8
Using the kinematics equation ::
d = Vi t + (0.5) a t2
- 30 = (65 Sin) (270 / ( 65 cos)) + (0.5) (-9.8) (270 / ( 65 cos))2
-30 = 270 tan - 84.55 (1 + tan2 )
solving the quadratic equation , we get
= 12.22 and 71.45
at = 12.22 , the ball hits the mountain, so we neglect it
at = 71.45 , the ball does not hit the mountain, so this is the right angle.
Along the y-direction , P to C motion :::
vertical distance = d = - 100 m
acceleration = a = g = -9.8
Using the kinematics equation ::
d = Vi t + (0.5) a t2
- 100 = (65 Sin) (270 / ( 65 cos)) + (0.5) (-9.8) (270 / ( 65 cos))2
-100 = 270 tan - 84.55 (1 + tan2 )
solving the quadratic equation , we get
= 72.895
at = 71.45 , the ball does not hit the
mountain, so this is the right angle.