Question

103. A body starts from rest and falls vertically
through a height of 20 m. If g = 10 m/s², the time
taken by the body to fall through the last 2
metres of the fall is :
(a) 1.9 s
(b) 0.1 s
(c) 0.01 s
(d) 2 s
20% 2-1​

Answer 1

Answer:

(b) 0.1 s

Explanation:

"The time taken by the body to fall through the last 2

metres" this means time taken by the body to travel from 18 m to 20 m and let it be t.

We will approach in this manner,

At first, we will find the velocity of that body at 18 m.

Second, we will find the time taken for the body to travel from 18 m to 20 m.

At 18 m the body had fallen 18 m with initial velocity 0 m/s and acceleration, g= 10 m/s² and let the velocity at 18 m be v.

Using the formula $v^{2} -u^{2} =2as$ ( Here v is final velocity, u is initial velocity , a is acceleration and s is distance covered by the body),

v²-0² = 2 x 10 x 18 ( Here I took downward direction as positive direction)

⇒v² = 360

⇒v  =√360

⇒v  = 6√10 m/s

Now take the moment in which the body falls 2 m ( 18 m to 20 m).

Here initial velocity is 6√10 m/s , acceleration is due to gravity (10 m/s²),

distance covered is 2 m.

Using the formula  $s=ut+\frac{1}{2} at^{2}$( here s is distance covered in time t , u is initial velocity, a is acceleration),

2=6√10t +1/2 x 10 x t²

⇒5t²+6√10-2=0, this is a quadratic equation in t.

using the quadratic formula -b±√(b²-4ac)/2a

t = (-6√10±√360 - (-40)) /2 x 5

⇒t  = -6√10±20÷10

Here minus sign will not work because time can't be negative

therefore t = -6√10 + 20 ÷ 10

-6√10 + 20 is roughly 1.02

hence, t = 1.02/10

Finally t = 0.102≈0.1 s