Projectile motion boat gun problem

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SUMMARY

The projectile motion problem involves a destroyer's 5-inch gun firing a projectile that travels 23,500 meters in 135 seconds. The solution requires calculating the initial velocity by determining both the vertical and horizontal components of the projectile's motion. The projectile spends 67.5 seconds ascending and 67.5 seconds descending, allowing for the calculation of the vertical component using the equation for free fall. The horizontal component can be derived from the total distance traveled, and the initial velocity can be found using the Pythagorean theorem.

PREREQUISITES
  • Understanding of kinematic equations, specifically v=vi+at and s=vi*t+.5at^2
  • Knowledge of projectile motion concepts, including vertical and horizontal components
  • Familiarity with Pythagorean theorem for calculating resultant velocities
  • Basic trigonometry for angle calculations and coordinate conversions
NEXT STEPS
  • Calculate the vertical component of initial velocity using free fall equations
  • Determine the horizontal velocity component based on total distance and time
  • Apply the Pythagorean theorem to find the resultant initial velocity
  • Explore Rectangular-to-Polar coordinate conversions for angle determination
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators seeking to explain these concepts in practical scenarios.

jesusu2
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Homework Statement


A speedy destroyer’s 5 inch gun fires a projectile at some angle to the horizontal. If the thing travels a distance of 23 500 m in 135 s, what was the projectile’s initial velocity?


Homework Equations


v=vi+at
v^2=vi^2+2a(s) s=displacement
s=vi*t+.5at^2


The Attempt at a Solution


Hello, I really don't know where to start, since I don't have an angle. I've made a diagram and split into 3 parts, up, down, and horizontal. How do I progress through this problem!?
 
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jesusu2 said:

Homework Statement


A speedy destroyer’s 5 inch gun fires a projectile at some angle to the horizontal. If the thing travels a distance of 23 500 m in 135 s, what was the projectile’s initial velocity?


Homework Equations


v=vi+at
v^2=vi^2+2a(s) s=displacement
s=vi*t+.5at^2


The Attempt at a Solution


Hello, I really don't know where to start, since I don't have an angle. I've made a diagram and split into 3 parts, up, down, and horizontal. How do I progress through this problem!?

Since this was a destroyer, the shell was fired from sea level, and lands at sea level so at least it lands at the same height from which it started.

The shell got to its target after 135s. That means it spent 67.5 seconds going up, then another 67.5 seconds coming back down.
For something to do that, you can calculate the vertical component of the initial velocity. [ie how fast is an object traveling if it falls for 67.5 seconds]

It landed 23 500m away after 135 seconds, so you can work out its horizontal velocity component.

One you have those two it is off to Pythagoras [and trigonometry if you also want the angle].
Note: you can use Rectangular-to-Polar co-ordinate conversions on you calculator if you are clever enough.
 

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