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# Jack Sparrow Problem

## Question

I have found the horizontal velocity and distance and the vertical acceleration but I can't find much else. If anybody else has done this problem on the worksheet please help!!

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2. Captain Jack Sparrow launches a cannonball from the deck of “The Black Pearl” (assume ground level) toward Davey Jones, sitting on a beach exactly 1 km away drinking his bottle of rum completely unaware of the impending attack. If the cannon ball is launched at an angle of 33 degrees with the horizontal at a velocity of 103 m/s, how close will it get to hitting Davey Jones? How long will it stay in the air? What is the cannonball’s maximum height?

[ATTACH=CONFIG]39[/ATTACH] First you must break up the initial velocity of 103 m/s at an angle of 33 degrees into its components:

$\ {v_{ix}} = {v_i}\cos \theta = (103{\textstyle{m \over s}})(\cos {33^ \circ }) = 86.4{\textstyle{m \over s}}$

$\ {v_{iy}} = {v_i}\sin \theta = (103{\textstyle{m \over s}})(\sin {33^ \circ }) = 56.1{\textstyle{m \over s}}$

Next, we can set up our horizontal and vertical tables for motion.

Horz:Vert:

vi = 86.4 m/s

vf = 86.4 m/s

d = ?

a = 0

t = ?

Vert:

vi = 56.1 m/s

vf = ?

d = ?

a = -9.81 m/s^2

t = ?

To solve this, I need to know one more variable in the vertical direction. Let's say I cut the motion of the cannonball in half. At its highest point, halfway through its motion, it's vertical velocity is 0. Now, my vertical table (for half the motion) looks like this:

Vert:

vi = 56.1 m/s

vf = 0

d = ?

a = -9.81 m/s^2

t = ?

From here, I can solve my vertical table for the time that the cannonball is in the air on the way up. To get the total time in the air, I just double my answer.

To get the Cannonball's maximum height, solve for d in the vertical direction.

To get the Cannonball's range, take the TOTAL time in the air and solve for the horizontal displacement. Use this to tell you how close the cannonball gets to hitting Davey Jones.

Good luck!

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