Projectile motion astronaut problem

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An astronaut on the moon strikes a golf ball at a velocity of 32 m/s at an angle of 35 degrees, with the moon's gravitational acceleration at 1.6 m/s². The vertical component of the ball's velocity is calculated to be 18 m/s. Using the kinematic equation, the time of flight is determined to be approximately 22.5 seconds. The calculations show that the maximum height reached by the ball is 189 meters before it lands in a crater 15 meters below the launch level. The discussion highlights the importance of correctly applying kinematic equations in projectile motion problems.
Morhas
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Homework Statement


An astronaut on the moon, where g=1.6m/s/s, strikes a golfball giving the ball a velocity of 32m/s [35degrees above the moons horizontal]. The ball lands in a crater floor that is 15m below the level where it was struck. Determine:

a) The maximum height of the ball


Homework Equations





The Attempt at a Solution



sin(35)=Vy/32
Vy= 18m/s

d=V1t+1/2at^2
0=18t + 0.5(-1.6)t^2
0.8t=18
t=22.5

d=18(22.5/2)+0.5(-1.6)(22.5/2)
d=198-8.8
d=189m
 
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Hi Morhas,

Morhas said:

Homework Statement


An astronaut on the moon, where g=1.6m/s/s, strikes a golfball giving the ball a velocity of 32m/s [35degrees above the moons horizontal]. The ball lands in a crater floor that is 15m below the level where it was struck. Determine:

a) The maximum height of the ball


Homework Equations





The Attempt at a Solution



sin(35)=Vy/32
Vy= 18m/s

d=V1t+1/2at^2
0=18t + 0.5(-1.6)t^2
0.8t=18
t=22.5

d=18(22.5/2)+0.5(-1.6)(22.5/2)

If you compare this to the equation you have above, you can see the time is supposed to be squared in the second term on the right hand side.

d=198-8.8
d=189m
 
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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