I,m confused, If I find the distance for the upward motion using delta v x t = delta d, where we can find the t by using formula t = delta v / g = 0.5 seconds, why did I get the upward distance= 2.45m? since the upward distance= (v final- v initial) x t= 4.9 m/s x 0.5s = 2.45 m? delta v is not negative since I assume that falling distance will be same as upward distance due to same upward and downward velocity. May I know why the distance is 1.225m instead of 2.45m?
Hi! Thanks for the video! I have a question, In my physics lab today we did a lab where we found and graphed data for a ball bouncing. One of the questions was to identify where the ball is in free fall and the region where it’s moving upwards and moving downwards in free fall. I had put it’s moving downwards at ball drop then upwards once it bounces BUT would free fall actually be right after the bounce (moving upwards then downwards..ie the parabola)? Since the ball bounces between the downward and upward for my original answer, is that still considered free fall? (Sorry if this isn’t very clear btw)
The ball would be in free fall anytime that it is in the air, doesn't matter whether it's moving up or down. The only time that the ball is NOT in free fall is when it is in contact with the ground. Hope this helps!
I have a similar problem, but the initial "Y" is 1.80 and since I need to solve for "time", the t would just be on both sides without cancelling one of them. I'm given Y (0). Yo(1.80m). g. Vo(4m/s)... that's it
the most perfect refresher i coulda found ever, ty fam gg wp keep grinding
Such a W video
Keep going man! ♥
Thanks, will do!
The front of the train gets to the platform 1st but stops at the same time as the rear carriage because of the connection
May i ask why use negatice sign on gravity?
I,m confused, If I find the distance for the upward motion using delta v x t = delta d, where we can find the t by using formula t = delta v / g = 0.5 seconds, why did I get the upward distance= 2.45m? since the upward distance= (v final- v initial) x t= 4.9 m/s x 0.5s = 2.45 m? delta v is not negative since I assume that falling distance will be same as upward distance due to same upward and downward velocity. May I know why the distance is 1.225m instead of 2.45m?
Thank youuuuuuu so much!!!
No prrooooblem!!!!
Thank you
You're welcome!
Hi! Thanks for the video! I have a question, In my physics lab today we did a lab where we found and graphed data for a ball bouncing. One of the questions was to identify where the ball is in free fall and the region where it’s moving upwards and moving downwards in free fall. I had put it’s moving downwards at ball drop then upwards once it bounces BUT would free fall actually be right after the bounce (moving upwards then downwards..ie the parabola)? Since the ball bounces between the downward and upward for my original answer, is that still considered free fall? (Sorry if this isn’t very clear btw)
The ball would be in free fall anytime that it is in the air, doesn't matter whether it's moving up or down. The only time that the ball is NOT in free fall is when it is in contact with the ground. Hope this helps!
Q
Do we ignore ball's mass or should it be counted to this equation?
You should ignore the ball's mass.
Would it be a free fall if I throw with force my initial velocity is 5m/s ?
NO
I have a similar problem, but the initial "Y" is 1.80 and since I need to solve for "time", the t would just be on both sides without cancelling one of them. I'm given Y (0). Yo(1.80m). g. Vo(4m/s)... that's it
Wait isn't displacement v final -v initial over time?
No, v final - v initial over time would be acceleration. Displacement is just x final - x initial.
If a object is thrown from certain height then what will be the acceleration due to gravity?
g is always 9.8 m/s/s as long as you ignore air resistance and are near the surface of the earth .
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