# Finding Final Velocity or Setting it Equal to 0(Kinematics)

Lori
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So, i have this sample problem to reference to :

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of the release and the instant of impact with the ground? Here neglect air resistance.

I know that I'm suppose to solve for time with the given information. I have intitial y velocity which is 20 m/s and i know the y(height) is 60. Should i assume that Vf is 0 because it hits the floor? And, i use the Vy = Vyi + gt to get time about 2 seconds?

Or, should i find final velocity with vy^2 = vyi^2 + 2g(y) and THEN use vy to solve for t in Vy = vi + gt?

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TSny
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<Moderator's note: Moved from a technical forum and thus no template.>
Should i assume that Vf is 0 because it hits the floor?
No. The final velocity refers to the final velocity of the projectile motion, just before striking the ground.

Or, should i find final velocity with vy^2 = vyi^2 + 2g(y) and THEN use vy to solve for t in Vy = vi + gt?
That should work. Be careful with signs.
Alternately, you can get the answer using just one appropriate equation.

Lori
No. The final velocity refers to the final velocity of the projectile motion, just before striking the ground. Since im examining the ball when it's going downward, g would be positive?

That should work. Be careful with signs.
Alternately, you can get the answer using just one appropriate equation.
Also, if i have to plug in 9.8 for acceleration , is it safe to say the g will be negative whenever the upward direction is positive and positive when upward is negative? In this example, downward is negative, so i would use positive g right?

TSny
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Also, if i have to plug in 9.8 for acceleration , is it safe to say the g will be negative whenever the upward direction is positive and positive when upward is negative? In this example, downward is negative, so i would use positive g right?
This causes a lot of confusion. Different instructors sometimes use different conventions here. So, I'm not sure what your instructor uses.

My preference is to always use the symbol g to represent the magntitude of the acceleration of gravity. So, g is always a positive number whether you take upward as positive direction or downward as positive direction. g = 9.8 m/s2 near the earth's surface.

The velocity equation for the y-direction is v2yf = v2yi + 2ayΔy.

You have a choice of taking the positive direction for y to be upward or downward. But you must make a choice and stick with it. If you take upward as positive, then the acceleration due to gravity is negative (since the acceleration is downward). Thus, ay = -g = -9.8 m/s2. The equation is then

v2yf = v2yi + 2ayΔy = v2yi + 2(-g)Δy

So,

v2yf = v2yi - 2gΔy where g = 9.8 m/s2

(But again, there are those who prefer to define g as a negative number for the case where upward is taken as positive. This is fine as long as it is treated properly when used in the equations.)

I think the safest thing to do is always write the equation as v2yf = v2yi + 2ayΔy . If you are dealing with a projectile problem where you choose upward as positive, then just make sure you plug in a negative number for ay.

Lori
This causes a lot of confusion. Different instructors sometimes use different conventions here. So, I'm not sure what your instructor uses.

My preference is to always use the symbol g to represent the magntitude of the acceleration of gravity. So, g is always a positive number whether you take upward as positive direction or downward as positive direction. g = 9.8 m/s2 near the earth's surface.

The velocity equation for the y-direction is v2yf = v2yi + 2ayΔy.

You have a choice of taking the positive direction for y to be upward or downward. But you must make a choice and stick with it. If you take upward as positive, then the acceleration due to gravity is negative (since the acceleration is downward). Thus, ay = -g = -9.8 m/s2. The equation is then

v2yf = v2yi + 2ayΔy = v2yi + 2(-g)Δy

So,

v2yf = v2yi - 2gΔy where g = 9.8 m/s2

(But again, there are those who prefer to define g as a negative number for the case where upward is taken as positive. This is fine as long as it is treated properly when used in the equations.)

I think the safest thing to do is always write the equation as v2yf = v2yi + 2ayΔy . If you are dealing with a projectile problem where you choose upward as positive, then just make sure you plug in a negative number for ay.
In this problem, I treated velocity as positive for the intial velocities even though the object was thrown downward. and i used acceleration positive g. So, what i am doing is that I made the positive direction downward, so i had to use a positive g?

And lets say the object is going up at 20m/s. I would treat the acceleration as a negative since I let the positive direction as upward direction

TSny
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In this problem, I treated velocity as positive for the intial velocities even though the object was thrown downward. and i used acceleration positive g. So, what i am doing is that I made the positive direction downward, so i had to use a positive g?
Yes. Except that I think it's better to say that you used a positive value for ay. As I said, different people adopt different conventions concerning the sign of the symbol g. But ay would definitely be positive for everybody when downward is considered the positive direction and when the acceleration is due to gravity.

And lets say the object is going up at 20m/s. I would treat the acceleration as a negative since I let the positive direction as upward direction
Yes, if you now take the positive direction as upward, ay would be a negative number since the direction of acceleration is downward. I would express this as ay = -g where g is a positive number as explained in my previous post. But, there are people who would express this as ay = g where g is a negative number. Either way, ay is negative and that's what matters when you take upward as positive.

• Lori