Question

ratio of the distances travelled by a freely falling body in first ,second, and third second of its fall

Answer 1

To find,

The ratio of ratio of the distances travelled by a freely falling body in first, second, and third second of its fall .

Solution,

The distance covered by an object during free fall is given by :

$d=ut+\dfrac{1}{2}at^2$

Here, u = 0, a = g

So,

$d\propto t^2$

For 1 second,

$d_1=kt^2=k$

For 2 second,

$d_2=k(2)^2=4k$

For 3 second,

$d_3=k(3)^2=9k$

Distance covered in 2 seconds, 4k-k = 3k

Distance covered in 3 seconds, 9k-4k = 5k

So, the ratio of distances traveled by a freely falling body in first, second, and third second will be 1:3:5.

Answer 2

ans is 1:3:5 this is answer of ur qsn

exp:To find,

The ratio of ratio of the distances travelled by a freely falling body in first, second, and third second of its fall .

Solution,

The distance covered by an object during free fall is given by :

d=ut+\dfrac{1}{2}at^2d=ut+

2

1

at

2

Here, u = 0, a = g

So,

d\propto t^2d∝t

2

For 1 second,

d_1=kt^2=kd

1

=kt

2

=k

For 2 second,

d_2=k(2)^2=4kd

2

=k(2)

2

=4k

For 3 second,

d_3=k(3)^2=9kd

3

=k(3)

2

=9k

Distance covered in 2 seconds, 4k-k = 3k

Distance covered in 3 seconds, 9k-4k = 5k

So, the ratio of distances traveled by a freely falling body in first, second, and third second will be 1:3:5.