Position of Ball w/ Initial Velocity 10 m/s After 2.12794 s

AI Thread Summary
The discussion focuses on calculating the position of a ball after 2.12794 seconds with an initial velocity of 10 m/s under the influence of gravity, which is 9.8 m/s². For a downward throw, the correct formula is y = Vot + 1/2gt², leading to a position calculation that factors in the negative acceleration due to gravity. For an upward throw, participants clarify the need to account for the downward direction of gravity in the calculations, emphasizing the importance of vector direction. The calculations provided by users indicate confusion over the signs and values, leading to incorrect results. Accurate application of the equations is crucial for determining the correct position of the ball.
kimikims
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A ball has an initial velocity of 10 m/s.
The acceleration of gravity is 9.8 m/s^2.

What will be it's postion after 2.12794 s if it is throw

a) down with an initial velocity of 10 m/s? Answer in units of m.

b) up with an initial velocity of 10 m/s? Answer in units of m.

--

I think for b you use:

y = Vot + 1/2gt^2

I got 43.47m, but that's wrong.
 
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I got -1.03, i think your calculations are wrong, and also i forgot if its negative it's pointing down, so it's going down.
 
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use this : y = 10t - 9.8t²/2 and the t is given

the g-vector points downwards the y-axis thus you need to write a minus sign. Watch out when you work with vectors : you have the magnitude of a vector AND the direction of a vector...

marlon
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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