1) A cannon is at a castle on a cliff and at a hight of 200 m above the sea level. Thecannon barrel points at an angle of +30.0◦from the horizontal (pointing slightly upwards)and can give a maximum speed to the cannon ball of 70.0 m/s. Ignoring air resistancehow close in the horizontal displacement should a ship be in order for the cannon to beable to hit the ship. (hints: identify the motion, write the equations and calculate firstthe total time the ball is in the air). A ship has some size so we do not need an accuracybetter than 10 m.a. about 6.6 ×102 mb. about 5.3 ×102 mc. about 4.3 ×102 md. about 2.2 ×102 m

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Asked Sep 25, 2019
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1) A cannon is at a castle on a cliff and at a hight of 200 m above the sea level. The
cannon barrel points at an angle of +30.0◦
from the horizontal (pointing slightly upwards)
and can give a maximum speed to the cannon ball of 70.0 m/s. Ignoring air resistance
how close in the horizontal displacement should a ship be in order for the cannon to be
able to hit the ship. (hints: identify the motion, write the equations and calculate first
the total time the ball is in the air). A ship has some size so we do not need an accuracy
better than 10 m.
a. about 6.6 ×102 m
b. about 5.3 ×102 m
c. about 4.3 ×102 m
d. about 2.2 ×102 m

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Expert Answer

Step 1

The above problem is an example of projectile motion.

The velocity of projection u can be resolved into horizontal and vertical components as,

uucos30°
usin 30°
и,
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uucos30° usin 30° и,

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Step 2

Substitute 70 m/s for u in the above two expressions and solve for ux and uy as,

70cos 30°
= 60.62m/s
u = 70sin 30°
35m/s
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70cos 30° = 60.62m/s u = 70sin 30° 35m/s

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Step 3

Using the second kinematic equation for motion of the ...

1
y= ut-
gt
2
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1 y= ut- gt 2

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