Projectile motion with a few variables

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The discussion focuses on solving a projectile motion problem involving uniform acceleration, where a projectile is launched from a height of 1 meter at a 60-degree angle and lands 60 meters away. The initial attempts to use standard equations for vertical and horizontal motion resulted in complex equations that were difficult to solve. Participants emphasized the need to utilize basic projectile motion equations and the known sine and cosine values for the launch angle to simplify calculations. After some guidance, the original poster realized their misunderstanding and expressed gratitude for the assistance. The conversation highlights the importance of correctly applying projectile motion principles to solve such problems effectively.
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Homework Statement

Assume uniform acceleration. The projectile is released at a height 1m above the ground traveling at angle 60 above the horizontal. It hits the ground 60m away from the launch position. Calculate acceleration of the particle in acceleration phase.[/B]

Homework Equations


s=ut+1/2ut^2 in y direction
s=ut in x direction[/B]

The Attempt at a Solution


tried using the above two and substituting for time and initial velocity but both give horrible equations which i can't solve.
Tried the range formula but that doesn't work because of the 2m start position.[/B]
 
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The simple equal height for launch and landing range formula is out, so you need to fall back on the basic projectile motion equations. Fortunately the angle provided has well known values for sine and cosine, so writing the projectile motion equations in terms of an unknown velocity v is made a bit easier.

Why don't you show us your attempt using the projectile motion equations and point out where you get stuck?
 
Got it now. I was just being silly thank you for your help :)
 
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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