Calculating Projectile Motion: Distance of Cannonball to Black Pearl

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SUMMARY

The discussion centers on calculating the horizontal distance from a cliff to the Black Pearl ship, given that a cannonball is fired at 50 m/s at an angle of 30° from a height of 100 m. The vertical displacement (Delta y) is -100 m, indicating the cannonball descends to sea level. The horizontal (Vx) and vertical (Vy) components of the initial velocity are calculated using the equations Vx = v cos(theta) and Vy = V sin(theta). These calculations are essential for determining the exact distance to the ship.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Basic knowledge of kinematic equations
  • Ability to perform vector decomposition
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  • Calculate the time of flight using the vertical motion equations
  • Determine the horizontal distance using the horizontal velocity and time of flight
  • Explore the effects of different launch angles on projectile distance
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SherazSiddiqu
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Captain Jack Sparrow shoots a cannonball at the black pearl which is located off the cliff that is 100m high. The cannon ball leaves the cannon at 50m/s at 30° to the horizon and hits the Black Pear(the ship) directly.
How far is the Black Pearl(the ship) from the cliff?

Please help thanks.
 
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Here's a picture:
35bvxab.png


So from that you can use this equation to solve:
l_vertical_displacement_equation.png

Delta y is the change in the y which is y2-y1 meaning the change in y will be -100 (NEGATIVE)Edit: I forgot you need to find the x and y components:
v cos (theta) = Vx
V sin (theta) = Vy - You need this 1, other is just for extra practice or w/e. (for this problem)
 

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