Finding Time For Projectile Motion

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To find the range of a marble launched at 6.89 m/s at a 15-degree angle, first determine the initial vertical and horizontal velocities using trigonometric functions. The vertical motion can be analyzed to find the time of flight, which is influenced by gravity. The horizontal velocity remains constant, allowing the range to be calculated using the formula distance equals velocity multiplied by time. It's important to remember that the total time of flight includes both the ascent and descent of the projectile. Completing these calculations will enable the determination of the marble's range.
phhysicssai
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Homework Statement



A marble launcher shoots a marble with a launch velocity of 6.89 m/s @15.0 degrees above horizontal.

-Find the marble's range.



(I tried to do this but I need to be able to find the time for horizontal motion... how would I do that?)

Homework Equations





The Attempt at a Solution

 
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phhysicssai said:

Homework Statement



A marble launcher shoots a marble with a launch velocity of 6.89 m/s @15.0 degrees above horizontal.

-Find the marble's range.



(I tried to do this but I need to be able to find the time for horizontal motion... how would I do that?)

Homework Equations





The Attempt at a Solution


The time is calculated by considering the vertical component of the motion, and treating it like any other vertical motion example.

Or you could have a read of the

http://en.wikipedia.org/wiki/Range_of_a_projectile

then you only need a substitution - provided your projectile is landing at the same level it was launched from and not from/onto a hill/cliff/building.
 
phhysicssai said:

Homework Statement



A marble launcher shoots a marble with a launch velocity of 6.89 m/s @15.0 degrees above horizontal.

-Find the marble's range.



(I tried to do this but I need to be able to find the time for horizontal motion... how would I do that?)

Because its 6.89m/s @ 15.0deg, you can use trigonometric functions to find the initial vertical and horizontal speed:

vertical:
sin(15.0) = o / 6.89

horizontal:
cos(15.0) = a / 6.89

Since the horizontal velocity is constant as gravity does not affect it, the time the projectile is in the air can be found by:

v = a*t

where

v = vertical
a = acceleration due to gravity
t = time of projectile

Once you have the time of the projectile its simple.

We know that distance = velocity * time

-> d = v * t

where d is the "range" we are trying to solve for
where v is the HORIZONTAL velocity
where t is the time of the projectile


Once you have done this, post your work/solution so I can help you further. Nobody here is going to do you homework for you, we can only HELP you if you are trying.
 
jack343, you are forgetting that the projectile goes up AND down so the time of flight is double what you calculate.
 
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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