Projectile Height Calculation: Baseball Thrown at 30 Degrees

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A baseball is thrown at an initial velocity of 20 m/s at a 30-degree angle, and the calculation for its maximum height involves using the equation 0 = Vi^2 + 2ad. The initial attempt yielded a height of 5.0 meters, but it was suggested to add the 2-meter release height, resulting in a total height of 7.0 meters. There was also a discussion about rounding and the use of gravitational acceleration, with some calculations yielding 5.10 meters, leading to a total height of 7.1 meters. The variability in gravitational acceleration used by the teacher was noted humorously. Accurate height calculations must consider both the projectile's motion and the initial height from which it is thrown.
bionut
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Homework Statement



A baseball is thrown with an initial velocity of 20 ms at an angle of 30 degrees with the horizontal. How high did it go if it was released and caught 2 meters from the ground?


Homework Equations





The Attempt at a Solution



0=Vi^2 + 2ad
0=(20Sin30)2 + 2 X (-9.8)d
19.62d=100
d=100/19.62
d=5.0m, but this is wrong... Would I mabey need to add 2m to my answer beacuse it is projected from a 2m height? therefore it will go 7m??
 
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bionut said:

Homework Statement



A baseball is thrown with an initial velocity of 20 ms at an angle of 30 degrees with the horizontal. How high did it go if it was released and caught 2 meters from the ground?


Homework Equations


The Attempt at a Solution



0=Vi^2 + 2ad
0=(20Sin30)2 + 2 X (-9.8)d
19.62d=100
d=100/19.62
d=5.0m, but this is wrong... Would I mabey need to add 2m to my answer beacuse it is projected from a 2m height? therefore it will go 7m??

Adding the 2 would be a very good idea , yes

EDIT: you did round off quite a bit to get 5.0m too. That is the answer you get if you take g = 10
 
I know... I got 5.10m, whihc would give 7.1m... I think my teacher sometimes uses 10 or 9.81 depends what kind of day it is ;-), lol
 
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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