Projectile motion boat gun problem

  • Thread starter jesusu2
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  • #1
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!?
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,435
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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!?

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 travelling 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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