Find if the boat will see the flare, what angle will they?

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SUMMARY

This discussion focuses on calculating the range of a flare fired from a deserted island to determine if a ship 300 meters offshore can see it. Using the range formula R = (vi^2 * sin(2 * theta)) / g, the initial velocity of 161 km/hr (44.72 m/s) at an angle of 33.3 degrees yields a range of approximately 187 meters, which is insufficient for visibility. To ensure the flare is seen, a firing angle of 45 degrees is recommended, resulting in a maximum range of around 200 meters, allowing the flare to be visible to the ship.

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Homework Statement


You have been stranded on a deserted island. Fortunately, you managed to bring a flare gun. You can see a ship 300. m off shore. Unfortunately, because of the daylight the ship cannot see your flare unless it is within 100. m of the ship.
A.) If you fire the flare gun from the shore line with an initial velocity of 161 km/hr at an angle of 33.3 degrees , will they see your flare? (You must show your work in explaining why/why not.)
B.) With this same initial speed, is there an angle you could fire that would allow them to see your flare? If so, what is that angle?

Homework Equations


R= (vi^2sin2(theta))/g

The Attempt at a Solution


plugging in the known information i get R=((44.72m/s)^2(sin(66.6))/9.8m/s^2 =187.

would this be correct? or is calculating the range just the x distance?

and i know 45 degrees is where the max is so for part b) i would plug in 45 degrees into the formula to get around 200. and therefore they'd see the flare.

When i initially did this problem i used the formula v^2=u^2+2as
u=44.72 m/s (sin(33.3)=24.6
a=-9.8 m/s^2
plugging in i got

0=(25.6^2)+2(-9.8)s
s=30.76
and this got me a distance of 30.76 (which is COMPLETELY different than the other formula. what is the correct way?

okay, so relooking at this i see that using range is how FAR it goes (horizontally) and the second way i showed is actually showing the max height it hits. so since I'm just looking for the flare to get at least 200.m off shore I'm assuming range is therefore the correct approach.

Would this be correct to think?
 
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u*sin(33.3) will give the vertical velocity and distance not the horizontal range
 
Parixit said:
u*sin(33.3) will give the vertical velocity and distance not the horizontal range
okay, that's what i thought, therefore i would want to use the range formula R=((44.72m/s)^2(sin(66.6))/9.8m/s^2 =187. because this will give the horizontal distance which is all this question is really wanting i think, because it doesn't tell me about how high the flare needs to be, just how close to the boat it needs to be.
 
Charlene said:
okay, that's what i thought, therefore i would want to use the range formula R=((44.72m/s)^2(sin(66.6))/9.8m/s^2 =187. because this will give the horizontal distance which is all this question is really wanting i think, because it doesn't tell me about how high the flare needs to be, just how close to the boat it needs to be.
Yes exactly. And for the second part you can find the angle taking S as 200 in the same range formula and finding the angle. You may get a range of angles with 45 deg as the mid point of that range
 
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