Two balls are dropped from same height at 1 second interval of time. The separation between the two balls after 3 seconds of the drop of first ball is?

smm
24 Points
9 years ago
the answer is 24.5metres . use the formula s=ut+1/2at2
ANANYA VERMA
11 Points
8 years ago
kindly give the elaborate solution to it.
what should be t , u etc???
basically what value we should take for t???

Mayank Saxena
23 Points
7 years ago
s is distance t is time and a is acceleration and u is velocity but in this situation we are going to use h=ut-1/2gt2 because we are dealing with vertically upward movement we will take g as 10m/sec2
Chinmayee Dash
13 Points
5 years ago
Distance travelled by the 1st ball after 3 sec
S1= 1/2gt^2
=1/2*10*9
=45m
Distance travelled by the 2nd ball after 2sec
S2= 1/2gt^2
=1/2*10*4
=20m
Distance between the balls after the 1st ball travel for 3sec
D=S1-S2
=45m-20m
=25m
34 Points
5 years ago
Given  TWO BODIES X,Y ARE dropped   downward from same height
FOR FIRST BODY , AFTER 3 SEC
U=0
g= 10 m/s2
S= ut+1/2 at
S= 0 × 3 + 1/2 × 10 × (3)
S1= 45m ..........(1)
For second After one min it is dropped
So
S2= 1/2 (10)(2)2
S2 = 20m
Separation = S1 - S2
45m - 20 m
25 m

Hope u understand  this and have a nice day