Don't Drop Your Camera 5.0 Seconds After Liftoff

[@sciencemysteries7480]

Hey Mr. P! What was the reason you didn't add the acceleration vectors for parts u and both

[@joanferreira3896]

Thank you again Prof.

[@alyssamenezes95]

I have a question. Wouldn't the acceleration of part two u change because you dropped the camera and therefore decreased in mass? The mass total would be lighter and therefore the acceleration would increase because the force from the jetpacks stayed constant. Or in this problem are we assuming that the camera has no mass? Thank you so much for the amazing videos though!!!

[@joudyrehan3640]

The acceleration is independant of mass! During free fall, mass is irrelevant

[@surajkrojha1342]

for part 2 camera BO said 52.5+(-9.81)(5) and mr.P wrote 52.5+(-9.81)(5)^2 o_O!! WELL IS IT A MISTAKE OR I M SOMEWHERE WRONG??? equation is also diffrent

[@FlippingPhysics]

suraj kr ojha You are correct that there is a mistake @ 7:26, however, you have the mistake reversed. The typo is on the board. The change in time is not squared in the equation. Sorry for the confusion and thanks for letting me know. I have added an annotation to point out the error.

[@science1878]

why does the second(last) answer have only 2 sig figs

[@Charlessen]

This video is "flipping" cool

[@FlippingPhysics]

HumdrumCheetah7 Thanks! :)

[@shauryasingh9553]

@Flipping Physics Why is the acceleration in the first part of the problem taken as 10.5 m/s^2? The jetpack has an acceleration of 10.5m/s^2 but at the same time the gravitational acceleration is 9.81 m/s^2 so shouldn't the net acceleration be 10.5-9.81 = 0.69m/s^2

[@parvinder9373]

Why initial velocity is taken zero

[@shaima3445]

the initial velocity of the object before moving(or at start) is taken as zero since it has zero kinetic energy. When the body is at rest and it starts moving. initial velocity is equal to zero(u=0),only when the object starts from rest.