Projectile Motion Calculations: Ship's Shot at an Enemy Ship and Island Peak

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SUMMARY

The discussion focuses on calculating the trajectory of a projectile fired from a ship towards an enemy ship, considering the height of an island peak. The ship is positioned 2,500 meters from the peak, which is 1,800 meters high, and fires a projectile with an initial velocity of 250 m/s at a 75-degree angle. The calculations reveal that the projectile lands 8 meters away from the enemy ship, which is located 610 meters beyond the peak, and reaches a vertical height of 210 meters relative to the peak. The relevant equations for vertical and horizontal motion of projectiles are provided for clarity.

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  • Understanding of projectile motion equations
  • Knowledge of trigonometric functions for angle calculations
  • Familiarity with kinematic equations in physics
  • Basic grasp of vertical and horizontal displacement concepts
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  • Learn how to apply kinematic equations to solve real-world problems
  • Explore the effects of different launch angles on projectile range
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lucianman24
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Please help me with the following question:

A ship maneuvers to within 2.50 x 10 ^ 3 m of an island's 1.80 x 10 ^ 3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10 ^ 2 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 2.50 x 10 ^ 2 m/s at an angle of 75 degrees, how close to the enemy ship does the projectile land? How close (vertically) does the projectile come to the peak? My physics book says the answer will be 8m and 210m but I want to know how to do it.

Vertical Motion of a Projectile that falls from rest
Vy,f = -gt (where g = a = 9.81) (t = time)
Vy,f^2 = -2gy (y = delta y= displacement in the y direction)
delta y = -1/2g(t)^2

Horizontal Motion of a Projectile
Vx = Vx,i = constant ( i = intial velocity)
delta x = Vxt ( delta x = dispalcement in the x direction)

Projectiles Launched At An Angle
Vx = Vi (cos theta) = constant (theta = degrees of angle)
delta x = Vi (cos theta)t
Vy,f = Vi (sin theta) - gt
Vy,f^2 = Vi^2 (sin theta)^2 - 2g(delta y)
delta y = Vi (sin theta)t - 1/2g(t)^2

Hope you can understand if not reply saying what you dont
Thank You
 
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lucianman24 said:
Please help me with the following question:

A ship maneuvers to within 2.50 x 10 ^ 3 m of an island's 1.80 x 10 ^ 3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10 ^ 2 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 2.50 x 10 ^ 2 m/s at an angle of 75 degrees, how close to the enemy ship does the projectile land? How close (vertically) does the projectile come to the peak?


THANK YOU

Hey, lucianman, welcome to the site!

What's your work so far for this problem? What values do you know? What equations might be useful?
 
lucianman24 said:
Please help me with the following question:

A ship maneuvers to within 2.50 x 10 ^ 3 m of an island's 1.80 x 10 ^ 3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10 ^ 2 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 2.50 x 10 ^ 2 m/s at an angle of 75 degrees, how close to the enemy ship does the projectile land? How close (vertically) does the projectile come to the peak?

THANK YOU

Show us some work. You should start off by writing down all the equations related to projectile motion (i.e. velocity, displacement, etc.).

P.S. Typing 'projectile motion' into the search box should be very useful too, since it is a highly frequent topic. :smile:
 

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