# Homework Help: Word problem

1. Sep 16, 2011

### Miike012

my work is inside the picture...

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2. Sep 16, 2011

### ehild

You need y(initial). So leave it on the right hand side of the equation and move everything else to the other side.

ehild

3. Sep 16, 2011

### Miike012

Wont that give me the same result?

4. Sep 16, 2011

### ehild

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

ehild

5. Sep 16, 2011

### Miike012

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

6. Sep 16, 2011

### Miike012

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

7. Sep 16, 2011

### Miike012

shouldnt it be negative instead of pos?

8. Sep 16, 2011

### vela

Staff Emeritus
By using a=-g, you chose the upward direction to be positive. The ball is thrown downward, so what should the sign of vinit be?

9. Sep 16, 2011

### ehild

Your original equation is y=y(init)+v(init)t-(1/2) g t2
At the final position, y=0. g=9.8 m/s2, v(init)=-5 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. Sep 16, 2011

### Miike012

When working with free falling objects... In general... Is acceleration always calculated to be -9.80 in the equation?????

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. Sep 16, 2011

### ehild

You can choose positive and negative direction arbitrary, but once decided, you must use it consequently.

g is posite, it means the magnitude of gravitational acceleration on the surface of the Earth. It is about 9.8 m/s2.

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/s2.

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/s2.

ehild

12. Sep 16, 2011

### Miike012

thankyou

13. Sep 16, 2011

### Miike012

Sorry.. Let me see if I understand...

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. Sep 16, 2011

### HallsofIvy

No, the acceleration due to gravity is always downward so if you take + in the upward direction, g is negative. It is the initial velocity that is positive if the object is thrown upward, negative if it is thrown downward, 0 if it is "dropped".

15. Sep 16, 2011

### Miike012

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

16. Sep 16, 2011

### Miike012

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

17. Sep 16, 2011

### vela

Staff Emeritus
Yes, the equation is written assuming the upward direction is positive.

The original equation you learn for constant acceleration a is$$y=y_0+{v_y}_0 t+\frac{1}{2}at^2$$If you take the upward direction to be positive, you have a=-g=-9.8 m/s2, so you get$$y=y_0+{v_y}_0 t-\frac{1}{2}gt^2$$This is the equation you're using, so by using it, you've assumed the upward direction is positive. On the other hand, if you take the downward direction to be positive, you would have a=+g=+9.8 m/s2, so you get$$y=y_0+{v_y}_0 t+\frac{1}{2}gt^2$$In both cases, g=+9.8 m/s2 since g is the magnitude of the acceleration due to gravity.

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. Sep 16, 2011

### ehild

Mike,

Suppose a body moves with uniform acceleration. Its position in terms of time is y(t)=y(init) + v(init) t+a/2 t2.
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/s2. 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/s2. So your equation is

y(t)=y(init)-5t-9.8/2 t2.

ehild

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