Question

Simple question:
A stone is dropped from building and 2 seconds later another stone is dropped. How far apart are these two stones by the time the first one has reached at a speed of 30 m/s.
(Take g=10m/s^2)

a) 80m

b) 100m

c) 60m

d) 40m

Answer 1

Well v=30m/s , u=0m/s , a=g=9.81m/s^2

Answer 2

Then calculate t

Answer 3

a=(v-u)/t

Answer 4

@AccessDenied @Algebraic! @Callisto @countonme123 @chandhuru @UnkleRhaukus @ajprincess @precal

Answer 5

calculate the time taken by the first stone to reach 30 and then add 2 seconds in next calculation of your height for your second stone..and you are done

Answer 6

@sauravshakya a=10m/s^2

Answer 7

a=g

Answer 8

60m or 40m?

Answer 9

yaa 40 m is answer but how?

Answer 10

i solved in mind :p @ghazi will explain

Answer 11

plz do it by step by step

Answer 12

first calculate t @mayankdevnani @DLS please share the kudos :)

Answer 13

you can use :
v=u+at
30=0+10t
t=3 seconds for first b

Answer 14

ok then

Answer 15

then for 2nd ball it..says..t+2 seconds right?
so t=5 seconds for 2nd ball..
find its speed

Answer 16

you have both the speeds omg
calculate distance o_O

Answer 17

ok then

Answer 18

\[t=\frac{ V }{ a }=\frac{ 30 }{ 10 }\] then for the second stone \[h= 0.5 *g*t^2 \] but here t= 3+2 now calculate h for both the stones and find the difference

Answer 19

that's what questionsays!!

Answer 20

so the 2nd ball was dropped two seconds later than the first but falls for 5 seconds, whereas the first ball fell for 3 seconds. makes sense.

Answer 21

ok then

Answer 22

i want asnwer

Answer 23

@mathslover

Answer 24

t=3 for first stone and t=3-2=1 for second stone

Answer 25

yup

Answer 26

A:
H=0.5*10*9
B:
H=0.5*10*1

45-5=40m :)

Answer 27

use \[H= 0.5*g*t^2\] for both the timings... @DLS did it