A Free-Fall Problem That You Must Split Into Two Parts

[@rishabtirupathi9431]

I love how Bo always seems uninterested in physics until he randomly comes up with the correct calculation inside his head.

[@Adumbb]

Mr. P, how come in this video we say velocity initial total and not just velocity initial? isn’t vi and vf instantaneous so we don’t have to say total

[@Adumbb]

does the velocity initial always equal the velocity final in a UAM problem like this?

[@moimzi1310]

Great Video! However, I have a question. At 2:25 you said that the height initial became change in height. But how could we have determined that when he had no idea what the final height was? In other words, why did the initial height turn into the change in height, when we only know the initial height, and not the final? Thanks

[@FlippingPhysics]

The final height is zero. The initial height is 124cm. The displacement in the y-direction = height final - height initial = 0 - 124cm = -124cm

[@Anveshkrishna]

Mr. P, why isnt the total initial velocity 0? I know I get the wrong answers, but you were holding the ball before throwing it, right? Does that mean velocity initial is the velocity of the ball right after it leaves your hand? Thanks❤

[@FlippingPhysics]

Please watch this video: www.flippingphysics.com/common-free-fall-pitfalls.html

[@zellah7888]

you are literally an angel, thank you so much for making these

[@FlippingPhysics]

You're so welcome!

[@longlostbuilder1856]

What I don't get is that you can actually apply the delta y final to find the velocity initial when the velocity fluctuates between the highest point and the lowest and still give you the right answer. My instinct would have been to try to find a way to solve for the top first or algebra the whole thing between the two stages.

[@renatoh.santosdasilva3080]

The velocity changes but the acceleration is constant, so there you go, you can solve the problem. But if the acceleration was a variable, we'd need Calculus.

[@shubhsharmaaa]

Good work

[@FlippingPhysics]

Thanks.

[@technotux7835]

I Don't get it, why change in total distance 1.24m it should be 2.48m+double the distance since the ball leave your hand up word and return again

[@carultch]

The total displacement is equal to 1.24 m in the downward direction. The total distance is a different concept, which is 1.0339 m + 1.0339 m + 1.24 m = 3.31 m. When velocity turns around, or changes direction, distance and displacement are going to have different magnitudes. The total displacement with negative defined as downward would be +1.0339 m - 1.0339 m - 1.24 m = -1.24m, assigning a sign to keep track of direction to each section of the motion as the terms are summed. The total distance assigns positive signs to each of these terms, ignoring direction, and adds up the distance traveled in each section to get 3.31 m.