Horizontally Launched Projectiles

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An airplane flying horizontally at 200 km/h needs to release flame retardant from an altitude of 180 m to effectively target a fire. The correct distance for the release is calculated to be 336 meters. The initial attempt at solving the problem misused the horizontal velocity as the vertical initial velocity, leading to incorrect time calculations. The key to solving this problem lies in correctly converting the horizontal speed from kilometers per hour to meters per second. Accurate unit conversion and understanding projectile motion principles are essential for determining the correct release distance.
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Homework Statement


An airplane loaded with flame retardant travels horizontally at 200. km/h. The altimeter indicates that the plane is 180. m above the ground. The pilot wants to know how far in advance of the fire to release the retardant. Please assist her.
The answer was given, which is 336 meters


Homework Equations


d=vit+1/2at^2


The Attempt at a Solution


So I plugged in all the numbers, 180 for d, 0 for vi, -9.81 for a. I got to 180=4.905t^2, which turns to 36.69. Then, I squared it, and got about 6.8, which is nowhere close to the time I need, which is 1.68. Please help me!
 
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This problem is similar to firing a bullet horizontally with velocity of 200m/s and an altitude of 180m and asking how far it will travel. In the equation you used the initial velocity in the vertical direction is zero but I think you substituted the initial horizontal velocity.

See horizontal projectile motion,

http://www.mrfizix.com/home/projectilemotion.htm

and,

https://www.google.com/search?tbm=i...w.,cf.osb&fp=d92e0af2da5034c4&biw=734&bih=439

Good luck.
 
Last edited by a moderator:
By Vi, I meant ViY. I don't understand what I did wrong from the pages you sent me to. I know I'm doing something way wrong, I'm just not sure what
 
corong1997 said:
By Vi, I meant ViY. I don't understand what I did wrong from the pages you sent me to. I know I'm doing something way wrong, I'm just not sure what

OK, your time looks right, so the distance should be t*200m/s which is much larger then the "right" answer. One of the numbers in your first post are wrong?
 
Yeah probably, I found these problems online, so they may be a bit messed up :mad:
 
Check your units. The problem specifies the horizontal velocity to be 200 km/h, not 200 m/s.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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