What Is the Angle of Initial Velocity for a Projectile at Half Maximum Height?

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To determine the angle of initial velocity for a projectile at half its maximum height, the projectile's speed at this point is three-fourths of its initial speed. The problem involves breaking down the initial velocity into horizontal and vertical components, using the equations of motion. The horizontal component remains constant, while the vertical component changes according to gravitational acceleration. Attempts to solve using the equations of motion have resulted in two unknowns, complicating the calculations. Ultimately, the challenge lies in accurately incorporating the given conditions into the equations to find the angle.
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Homework Statement



At half its maximum height, the speed of a projectile is three-fourth its initial speed. What is the angle of the initial velocity vector with respect to the horizontal


Homework Equations



any projectile equation


The Attempt at a Solution



?
 
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Split the total speed into a vertical and horizontal component.
u_horiz = V cos(angle) u_vertical = V sin(angle)
The horizontal component doesn't change and the vertical component obeys v^2 = u^2 - 2 g h
 
okay so i tried splitting the horizontal and vertical components, but in the end and it doesn't accomplish anything because you get 2 unknown variables, also how would you account for the givens, such as, at half the max height you have 3/4 the initial velocity?

also i have tried using the equation V^2=V(o)^2 + 2 a x
from that i set V to (3/4 V(o))
g to (9.81) and
x to (x/2)
in the end that doesn't work out either??
 
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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