Ophira’s insurance company has notified her that her premiums will increase due to a poor insurance score and a recent claim she
1 answer:
You might be interested in
Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s
the only force acting on the object, is gravity, using feet will then be -32ft/s²,
was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues
so, we'll do the integral to get v(t) then
when will it reach the ground level? let's set s(t) = 0
part B) check the picture below
the only force acting on the object, is gravity, using feet will then be -32ft/s²,
was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues
so, we'll do the integral to get v(t) then
when will it reach the ground level? let's set s(t) = 0
part B) check the picture below
Just download photomath bro, does everything for you.
