Question

the ratio of distance described by a body falling freely from rest in the last second of its motion to that in last but one second of its motion is 5:4 find the total time taken by the body to reach the ground​

Answer 1

Distance travelled in the n th second . N is proportion to ( 2n - 1 ).

Explanation:

If a body starts with velocity u, then distance traveled in last nth second,

dn = u+ a(2n-1). Here, a is uniform acceleration.

Distance traveled in (n-1) th second is

d(n-1)= u+a[2(n-1)-1]=u+a(2n-3)

For a freely falling body a=g=10m/s^2, and

u=0. Then,

dn/d(n-1)=(2n-1)/(2n-3)=5/4. Therefore,

4(2n-1)=5(2n-3). Then,

8n-4=10n-15 or

2n=11. Then,

n=5.5 s.

Answer 2

5.5

Explanation:

If a body starts with velocity u, then distance traveled in last nth second,

dn = u+ a(2n-1). Here, a is uniform acceleration.

Distance traveled in (n-1) th second is

d(n-1)= u+a[2(n-1)-1]=u+a(2n-3)

For a freely falling body a=g=10m/s^2, and

u=0. Then,

dn/d(n-1)=(2n-1)/(2n-3)=5/4. Therefore,

4(2n-1)=5(2n-3). Then,

8n-4=10n-15 or

2n=11. Then,

n=5.5 s.