Question

A body falls freely from rest, so that it covers as much distance in the last second of it's motion as covered in first four seconds. The body has fallen for a time

Answer 1

Given :

A body falls freely from rest, so that it covers as much distance in the last second of it's motion as covered in first four seconds

To Find :

Time taken to fall

Solution :

Distance travelled by the body in n seconds is ,

$\bigstar\ \; \sf \pink{s=ut+\dfrac{1}{2}at^2}$

where ,

• s denotes distance
• u denotes initial velocity
• a denotes acceleration
• t denotes time

Distance travelled by the body in nth second is ,

$\bigstar\ \; \sf \green{S_n=u+\dfrac{a}{2}(2n-1)}$

where ,

• Sₙ denotes distance in nth second
• u denotes initial velocity
• a denotes acceleration
• n denotes nth second

_________________________________

A body covers as much distance in the last second of it's motion as covered in first four seconds

Initial velocity , u = 0 m/s

∵ falls from rest

Acceleration , a = g = 10 m/s²

Distance covered in first 4 seconds is ,

$\to \sf s=ut+\dfrac{1}{2}at^2\\\\\to \sf s=(0)t+\dfrac{1}{2}g(4)^2\\\\\to \sf s=\dfrac{1}{2}(10)(16)\\\\\leadsto \sf \blue{s=80\ m}\ \; \bigstar$

Now , this distance is equals to the distance covered in nth ( last ) second ,

$\to \sf S_n=u+\dfrac{a}{2}(2n-1)\\\\\to \sf 80=(0)+\dfrac{g}{2}(2n-1)\\\\\to \sf 80=\dfrac{10}{2}(2n-1)\\\\\to \sf 80=5(2n-1)\\\\\to \sf 2n-1=16\\\\\to \sf 2n=17\\\\ \sf \red{\leadsto n=8.5\ s}\ \; \bigstar$

Answer 2

Answer:

Given :

A body falls freely from rest, so that it covers as much distance in the last second of it's motion as covered in first four seconds

To Find :

Time taken to fall

Solution :

Distance travelled by the body in n seconds is ,

\bigstar\ \; \sf \pink{s=ut+\dfrac{1}{2}at^2}★ s=ut+

2

1

at

2

where ,

s denotes distance

u denotes initial velocity

a denotes acceleration

t denotes time

Distance travelled by the body in nth second is ,

\bigstar\ \; \sf \green{S_n=u+\dfrac{a}{2}(2n-1)}★ S

n

=u+

2

a

(2n−1)

where ,

Sₙ denotes distance in nth second

u denotes initial velocity

a denotes acceleration

n denotes nth second

_________________________________

A body covers as much distance in the last second of it's motion as covered in first four seconds

Initial velocity , u = 0 m/s

∵ falls from rest

Acceleration , a = g = 10 m/s²

Distance covered in first 4 seconds is ,

\begin{gathered}\to \sf s=ut+\dfrac{1}{2}at^2\\\\\to \sf s=(0)t+\dfrac{1}{2}g(4)^2\\\\\to \sf s=\dfrac{1}{2}(10)(16)\\\\\leadsto \sf \blue{s=80\ m}\ \; \bigstar\end{gathered}

→s=ut+

2

1

at

2

→s=(0)t+

2

1

g(4)

2

→s=

2

1

(10)(16)

⇝s=80 m ★

Now , this distance is equals to the distance covered in nth ( last ) second ,

\begin{gathered}\to \sf S_n=u+\dfrac{a}{2}(2n-1)\\\\\to \sf 80=(0)+\dfrac{g}{2}(2n-1)\\\\\to \sf 80=\dfrac{10}{2}(2n-1)\\\\\to \sf 80=5(2n-1)\\\\\to \sf 2n-1=16\\\\\to \sf 2n=17\\\\ \sf \red{\leadsto n=8.5\ s}\ \; \bigstar\end{gathered}

→S

n

=u+

2

a

(2n−1)

→80=(0)+

2

g

(2n−1)

→80=

2

10

(2n−1)

→80=5(2n−1)

→2n−1=16

→2n=17

⇝n=8.5 s ★