What Is the Takeoff Speed of an Aircraft Based on a Watch's Angle?

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The discussion revolves around estimating the takeoff speed of an aircraft based on the angle of a watch held by a passenger during takeoff. The watch makes a 25° angle with the vertical, leading to the calculation of acceleration using the tangent function, resulting in an acceleration of approximately 4.57 m/s². By applying the formula for velocity, the estimated takeoff speed is calculated to be around 82.2 m/s. This speed translates to approximately 300 km/h, which aligns with realistic takeoff speeds for jetliners. The calculations and conclusions presented confirm the accuracy of the estimated takeoff speed.
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Francesca dangles her watch from a thin piece of string while the jetliner she is in takes off from JFK Airport. She notices that the string makes an angle of 25° with respect to the vertical while the aircraft accelerates for takeoff, which takes about 18s. Estimate the takeoff speed of the aircraft.

k so you know that the force of gravity is mg and it's going down, and that the acceleration is ma along the horizontal, and so can you take the tan(25)=a/9.8 to get an acceleration of 4.57m/s^2?

And so for velocity you would get V=Vo+at
V=0+(4.57)(18)
V=82.2m/s?

thanks for reading
 
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It seems to be correct. Which means the take-off speed is around 300 km/h which is realistic.
 
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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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