Question

find the ratio of distance travel in 1st sec 2nd sec and 3rd sec by a ball when fall from height of 1000 m
pls answer this
​

Answer 1

Answer:

Initial velocity of the body  u=0

Explanation:

∴  S

t

​

=0+

2

1

​

g(2t−1)=

2

1

​

g(2t−1)

Distance travelled in first second i.e. t=1,  S

1

​

=

2

1

​

g(2×1−1)=

2

1

​

g

Distance travelled in 2nd second i.e. t=2,  S

2

​

=

2

1

​

g(2×2−1)=

2

3

​

g

Distance travelled in third second i.e. t=3,  S

3

​

=

2

1

​

g(2×3−1)=

2

5

​

g

Answer 2

X = ut + 1/2 at^2
For 1st second -> 0x1 + 1/2g(1)2 = 5 m
For 2nd second -> 0x2 + 1/2g(2)2 = 20 m
For 3rd second -> 0x3 + 1/2g(3)2 = 45 m
Ratio = 5:20:45