- #1
Crackle
- 2
- 0
Ok guys.
I've just come back from Thanksgiving holidays and I just can't seem to grasp this concept.
I have an X, Y, and t from the experiment that we did in lab today. I have a Vy of 91.2cm. I now need to calculate a VyFinal ( Vyf ). From what I have gathered the equation should come out to :
Vyf = Vy + g * t
So, in that said, my equations for the four balls ( diferent sizes ) should come out to be :
Vyf = 91.2cm + ( -9.80m/s2 * .50s ) ( my time for the ball to drop from 91.2cm high )
This would leave me adding a (cm) and a (m/s). That doesn't seem right to me because the units are odd.
I dropped each ball ( tennis, golf, nerf, and bouncy ) three times and recorded the time as t, the distance from which the ball traveled away from the dropping point as X, and Y being my constant at 91.2cm high off the ground. ( I believe this is going to become negative because the ball is traveling downward )
I have to do calculations for Vyf and Vx.
If X = (Vx)(t), then Vx = X/t correct? If this is true then my Vx would then be the distance from the counter to where the ball dropped divided by the amount of time it took to hit the ground.
I think I am on the right direction, I just need some advice to get me going.
Thanks Everyone,
Crackle
I've just come back from Thanksgiving holidays and I just can't seem to grasp this concept.
I have an X, Y, and t from the experiment that we did in lab today. I have a Vy of 91.2cm. I now need to calculate a VyFinal ( Vyf ). From what I have gathered the equation should come out to :
Vyf = Vy + g * t
So, in that said, my equations for the four balls ( diferent sizes ) should come out to be :
Vyf = 91.2cm + ( -9.80m/s2 * .50s ) ( my time for the ball to drop from 91.2cm high )
This would leave me adding a (cm) and a (m/s). That doesn't seem right to me because the units are odd.
I dropped each ball ( tennis, golf, nerf, and bouncy ) three times and recorded the time as t, the distance from which the ball traveled away from the dropping point as X, and Y being my constant at 91.2cm high off the ground. ( I believe this is going to become negative because the ball is traveling downward )
I have to do calculations for Vyf and Vx.
If X = (Vx)(t), then Vx = X/t correct? If this is true then my Vx would then be the distance from the counter to where the ball dropped divided by the amount of time it took to hit the ground.
I think I am on the right direction, I just need some advice to get me going.
Thanks Everyone,
Crackle