Question

15"
9. A stone is dropped from a cliff at 2:30:30 p.m. (hour:
minute:second). Another stone is dropped from the
same point at 2:30:31 p.m. Find the separation
between the stones at (a) 2:30:31 p.m., (b) 2:30:35
p.m.
the
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Answer 1

Answer:

(a) 4.9 m

(b) 44.1 m

Explanation:

(a) Separation at 2:30:31

For the 1st stone:

Initial velocity (u) = 0m/s

time (t) = 1 sec

Acceleration due to gravity (g) = 9.8 m/s²

We know_

v = u+ gt

==> v = 9.8 × 1

==> v = 9.8 m/s

Now,

2gh = v²-u²

==> 19.8h = 96.04

==> h = 96.04/19.6

==> h = 4.9 m

For the 2nd stone

u = 0m/a

t = 0m/s

g = 9.8 m/s²

So, v = u+gt

v= 0+g×0

==> v =0 m/a

And , 2gh=v²-u²

==> 19.6h = 0

==> h = 0 m

Hence separation at 2:30:31 is — 4.9-0 = 4.9m

(b) Separation at 2:30:35

1st stone:

u = 0m/s

t = 5s

g = 9.8m/s²

Hence v = u+gt

==> v = 9.8×5

==> v = 49 m/s²

And , 2gh = v² - u²

==> 19.6h = 2401

==> h = 122.5 m

For the 2nd stone

u = 0m/s

t = 4s

g= 9.8m/s²

Hence v= u+gt

==> v= 9.8×4

==> v= 39.2m/s

Finally 2gh = v²-u²

==> 19.6h = 1536.64

==> h = 78.4 m

Hence separation at 2:30:35 is

122.5 - 78.4

= 44.1 m

Hence (a) = 4.9 m

(b) = 44.1 m

Hope it's correct:)