Solving for Velocity with Constant Acceleration Equation | Simple Question

[LogicX]

I am using an equation for constant acceleration to find velocity.

v^2=v(initial)^2+2a(x-x(initial))

Dropping a rock from 4 meters to the ground. Initial velocity is 0, a=9.8, and displacement is -4.

So v^2=2(9.8)(-4), and it SHOULD equal=-8.85m/s (I know this is the correct answer, I already got it right on my homework).

The velocity should be negative. But obviously my above answer is impossible because you can't take the square root of a negative. When using these constant acceleration equations I make the addition sign in the above equation negative if it is going upward (makes sense since gravity is working against it), and positive if it is going downward . Do I apply the negative sign after I solve, to show the direction? That just doesn't seem right.

Help!

 

[rock.freak667]

I am using an equation for constant acceleration to find velocity.

v^2=v(initial)^2+2a(x-x(initial))

Dropping a rock from 4 meters to the ground.

You are measuring with respect to the ground, initially it is 4m above ground so x₀=4. At the ground x=0, so you are finding 'v' at x=0.

So you'll get v²= 2(-9.81)(0-4)

 

[LogicX]

Why is 9.8 negative? And in the problem, the answer had to be negative in order to get the correct answer for a later part. Here, why don't I just post the whole thing:

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, what is its average acceleration during that contact?

The answer is 1.26x10^3m/s^2

So I got the answer I already posted for descent, 8.85, then figured out the velocity for ascent, which is 6.26. Then the average acceleration is change in velocity over the time it was in contact with the floor. In order to get the right answer 8.85 had to be negative so you could have:

(6.26+8.85)/.012=1.26x10^3. One of those velocities has to be negative, no matter how you cut it, and you have to take a square root to get both of them. So what is going on?

EDIT: Velocity is the derivative of position, and if the graph of the function is decreasing then the graph of its derivative is negative. So it WOULD seem that you apply the sign after you do the calculation, based on the fact that taking a square root of something means you end up with two answers, +/-.

 

[rock.freak667]

Why is 9.8 negative?

In the entire motion, you have both upward and downward motion. You needed to take a direction a positive. I took down as negative, so a=-9.81 m/s².

 

[rwisz]

I would like to commend you on correctly using the constant acceleration equation to solve for velocity. Your understanding of the concept is evident in your explanation.

In this scenario, the rock is being dropped from a height of 4 meters, meaning it is initially moving downward with an initial velocity of 0. As it falls, it experiences a constant acceleration due to gravity of 9.8 m/s^2.

When using the equation v^2 = v(initial)^2 + 2a(x-x(initial)), it is important to note the direction of motion. In this case, since the rock is moving downward, the displacement (x-x(initial)) is negative. This means that the final velocity (v) will also be negative, indicating that the rock is still moving downward with a velocity of -8.85 m/s at the moment it reaches the ground.

The negative sign in the final velocity represents the direction of motion, and it is correct to apply it after solving the equation. It is important to pay attention to the direction of motion when using these equations to ensure the correct interpretation of the results.

I hope this helps clarify any confusion you may have had. Keep up the good work in your studies of physics!