Easy physics question. Easy points!!!?

Question

An object fall from height h from rest and travels 0.5h in the last 1.00s. Find the time of its fall.

Thanks!

Answer 1

last second travel
d2 = 0.5*g*[ t^2 - (t-1)^2]= 0.5*g*[2t-1] = 0.5h

previously travelled distance (during t-1 s)
d1 = 0.5*g* (t-1)^2 = 0.5h

d1 = d2
0.5*g*[t^2 -2*t+1] = 0.5*g*[2t-1]
t^2 -4t + 2 = 0
t = 3.414 s
(other root is < 1, not feasible since fall lasted longer than 1 sec)

h = 57.18 m

Answer 2

nicely, photograph a soccer field with a tennis ball interior the midsection, and then on the tip of the sector there's a marble. there is often empty area between those 2 gadgets. Now make it into 3-D, with quite a few marbles orbiting around the tennis ball. that's additionally usually empty area. As you have gotten guessed the marble is representative of an electron, and the tennis ball a nucleus. So in case you scale it right down to the dimensions of an atom of Iron, and upload adequate at the same time so as that this is going to become a block of iron, then maximum of it would be empty area

Answer 3

e(t) = 1/2gt^2 =0.5h
e(t+1) =1/2g(t+1)^2 = h

1/2gt^2 +gt+1/2g =h
gt+1/2g= 0.5h and 1/2 gt^2= 0.5h
gt+1/2g = 1/2gt^2
gt^2-2gt-g=0 t^2-2t-1=0 t= ((2+sqrt8)2 =1+sqrt(2)
t+1= 2+sqrt2 s
Former answer took 0.5 h as 500m (wrongly)

Answer 4

Let 2h be the total drop distance.
Then 2h= 1/2(9.8) t(total)^2 = 4.9 t(total)^2
And h= 2.45 t(total)^2
But h= 1/2 (9.8) t(h)^2 =4.9 t(h)^2 So.....
t(total)^2=2* t(h)^2
And t(total)=t(h)+1.
You have 2 equations and 2 unknowns
Solve for t(total) and be sure you discard neg root.

Answer 5

thats not easy i dont know it..lol

Answer 6

3.o'clock??? i dunno