A stone is dropped from the roof of a high building. A second stone is dropped 3.00 s later. How far apart are

Question

A stone is dropped from the roof of a high building. A second stone is dropped 3.00 s later. How far apart are the stones when the second one has reached a speed of 13.0 m/s?
m

Answer 1

V=9.8t m/s
d=4.9t^2 m
2nd stone reaches 13.0 m/s in:
13.0=9.8t
t=13.0/9.8=1.33s
at that time it will have fallen 4.9*1.33^2=8.62 m

the 1st stone has been falling 4.33 s
in that time it will have fallen 4.9*4.33^2=91.72 m

the distance between them is 91.72-8.62=83.1 m

Answer 2

distance = speed * time
= 13.0 *3.00
=39.0m
Iam not sure my answer is correct. I think I find the distance of dropped second stone and the ground.

Answer 3

velocity = gravitational acceleration x time
v = gt
solving for time: t = v/g (g=9.8 meter per second squared)

when the second stone reaches 13.0 m/s is has been falling for:
t = 13.0 m/s / 9.8 meter per sec2
t = 1.3 seconds
The first stone has been falling for 3 additional seconds = 4.3 seconds

The difference in distance is given by

diff = 1/2 g (t1^2 - t2^2) = 1/2 9.8 (4.3^2 - 1.3^2)

= 4.9 (18.49 - 1.69) = 82.3 meters

Answer 4

I'm pretty sure you would need the weight of the stones in order to figure this out.

Answer 5

Haven't you left out some necessary stats to calculate this???