Help with physics plz !!! Quiz tomorrow.?

Question

A stone is dropped from the roof of a high building. A second stone is dropped 1.5s later. How far apart are the stones when the second one has reached a speed of 12m/s?

Plz help answer with steps included.
Thankyou.

Answer 1

For the second stone, we need to find out how long it takes for it to get to 12 m/s:

Vf = Vi + at
12 m/s = 0 m/s + (9.81 m/s^2)t
t = 1.22 s

We can also find out how far the stone has dropped at this point:

Vf^2 = Vi^2 + 2ad
(12 m/s)^2 = (0 m/s)^2 + 2(9.81 m/s^2)d
d = 7.34 m

Since the first stone had a 1.5 s head start, we know that when the second stone reached 12 m/s, the first one has been airborne for 2.72 s. We can use the same equations to find out how fast the stone was going at that point, and then how far it fell:

Vf = Vi + at
Vf = 0 m/s + (9.81 m/s^2)(2.72 s)
Vf = 26.68 m/s

Vf^2 = Vi^2 + 2ad
(26.68 m/s)^2 = (0 m/s)^2 + 2(9.81 m/s^2)d
d = 36.28 m

So the distance between the stones is then 36.28 m - 7.34 m = 28.94 m.

Answer 2

2) enable speed of lit be v=3 X 10^8m/s, d be distance=100m, time =t; a) nw u knw tht, distance=speed X time t=d/v t=a hundred/3X10^8 t=3.3 X 10^-5 sec b) comparable right here, jus enable speed of sound be v=333 m/s, d be distance=100m, time =t; lower back, t=d/v t=a hundred/333 t=0.3 sec! its become longr a million thts why i did it, simalalry u do the nxt a million! bst of success fo 2row!!