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- Thread starter Miike012
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- #2

ehild

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ehild

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Wont that give me the same result?

- #4

ehild

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You will not be confused with the lot of minuses. Try to type in, I can not read your file.

ehild

ehild

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-(5)(2) - (9.8/2)(4) = y(init).... same thing

- #6

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actually I get 29.6.... but its neg....

- #7

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shouldnt it be negative instead of pos?

- #8

vela

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ehild

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-(5)(2) - (9.8/2)(4) = y(init).... same thing

Your original equation is y=y(init)+v(init)t-(1/2) g t

At the final position, y=0. g=9.8 m/s

So the equation is

y(init)-5*2 -(9.81/2)(4)=0 or y(init)-[5*2 +(9.81/2)(4)]=0

You subtract (10+9.81*2) from y(init) and you get zero. What is y (init)?

When solving an equation, we isolate the unknown at one side, by adding everything else to both sides or subtracting everything else form both sides, or multiplying/dividing both sides...

Add 5*2+(9.81/2)(4) to both sides now

ehild

- #10

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because I origninally thought a = -g meaning -(1/2) (-g) t^2 = (1/2) (g) t^2

... but I see that the original equation already has a neg sign there for acc... "-" (1/2) g t^2.

- #11

ehild

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g is posite, it means the magnitude of gravitational acceleration on the surface of the Earth. It is about 9.8 m/s

If you choose that y is height then the upward direction is positive. Gravity acts downward, so the acceleration is -g or -9.8 m/s

If you choose the quantity y as the displacement in downward direction, then the downward acceleration is positive and it is g=9.8 m/s

ehild

- #12

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thankyou

- #13

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If distance is counted + in upward direction, then a = -g if an object is thrown upward....

If distance is counted + in upward direction, then a = g if an object is dropped from a building downward???

is this correct?

- #14

HallsofIvy

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g=9.8 m/s2, why did u not say -9.8 m/s^2?

- #16

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Is it because the - sign is already in the equation?

- #17

vela

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The original equation you learn for constant acceleration

As ehild said, both ways will work, but you have to be consistent and adjust the signs on the various quantities depending on which convention you're following.

- #18

ehild

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Mike,

Suppose a body moves with uniform acceleration. Its position in terms of time is y(t)=y(init) + v(init) t+a/2 t^{2}.

In case gravity is the only force acting on the body, you know that it accelerates downward, and the magnitude of the acceleration is g=9.8 m/s^{2}. The value of g can be given with more or less accuracy, and it is different at different places of Earth, but it is always a positive number.

Before you start to solve a problem you draw a figure and show what direction you mean positive. You should choose the sign of both the acceleration and initial velocity**with respect to this direction**.

In the problem, y(init) is positive, and the position of the stone is always positive till it reaches the ground, where y is defined zero.

The initial velocity is downward, so it is negative, -5 m/s. Have been the stone thrown upward, the initial velocity would be positive.

The acceleration is downward, and its magnitude is equal to g, so a=-g =-9.8 m/s^{2}. So your equation is

y(t)=y(init)-5t-9.8/2 t^{2}.

ehild

Suppose a body moves with uniform acceleration. Its position in terms of time is y(t)=y(init) + v(init) t+a/2 t

In case gravity is the only force acting on the body, you know that it accelerates downward, and the magnitude of the acceleration is g=9.8 m/s

Before you start to solve a problem you draw a figure and show what direction you mean positive. You should choose the sign of both the acceleration and initial velocity

In the problem, y(init) is positive, and the position of the stone is always positive till it reaches the ground, where y is defined zero.

The initial velocity is downward, so it is negative, -5 m/s. Have been the stone thrown upward, the initial velocity would be positive.

The acceleration is downward, and its magnitude is equal to g, so a=-g =-9.8 m/s

y(t)=y(init)-5t-9.8/2 t

ehild

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