Projectiles (delta)(theta) = g(delta)t/V0

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The discussion centers on the projectile motion equation (delta)(theta) = g(delta)t/V0, where V0 represents the initial speed of the projectile and g is the acceleration due to gravity. Participants emphasize the importance of providing complete problem statements and relevant equations when seeking help. The conversation highlights the necessity of showing prior work to facilitate effective tutoring and correction of mistakes. This approach ensures a deeper understanding of the physics involved in projectile motion.

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  • Familiarity with the equations of motion in physics
  • Knowledge of the acceleration due to gravity (g)
  • Ability to manipulate algebraic equations
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calgaryleb
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Projectiles...(delta)(theta) = g(delta)t/V0

I am given this problem...
if the projectile is fired a short time (delta)t later, show that the angle correction is
(delta)(theta) = g(delta)t/V0
where V0 is the inital speed of the projectile and by assuming that the elevations of the projectile and the target are about the same.
g is acceleration due to gravity


Any ideas on how to show/proof this concept..
any/all idea is appreticated :smile:
 
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if the projectile is fired a short time (delta)t later

A short time later than what? Is there more to this question? What have you tried so far?
 
Welcome to the PF, calgaryleb. One of the most important rules here on the PF is that we do not do your homework or coursework or test problems for you. We are happy to offer tutorial help, and refer you to sources of information, and correct mistakes we see in your work, but we will not give you answers.

So please list the relevant equations (there's a reason that those are asked for in the Homework Help Template that you deleted when you made this post...), and please show us your work so far. And as hage567 says, you need to list the full question as well (gee, that's in the Homework Help Template too...)
 
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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