- #1

- 83

- 0

Ok now i am on acceleration due to gravity questions and i am kind of stuck.

Anyways the question is below

Heres what i did.

d=Vit + 1/2(a)(t^2)

d= -2.1(3.8) + 1/2(9.8)(3.8)^2

I think this how you get the answer which is 63 m but i dont get why we are using 3.8 s because at 3.8 s the displacement or distance should be 0 so can anyone please help me out

Sorry im am such an idiot i did the wrong steps the actual question is in the last post

Anyways the question is below

Heres what i did.

d=Vit + 1/2(a)(t^2)

d= -2.1(3.8) + 1/2(9.8)(3.8)^2

I think this how you get the answer which is 63 m but i dont get why we are using 3.8 s because at 3.8 s the displacement or distance should be 0 so can anyone please help me out

Sorry im am such an idiot i did the wrong steps the actual question is in the last post

Last edited: