Question

A body is freely falling. The displacement in the last second is equal to the displacement in the first three
seconds. What is the time of fall?

Answer 1

QUESTION:-

 A body is freely falling. The displacement in the last second is equal to the displacement in the first three  seconds. What is the time of fall?

EXPLANATION:-

u=0 m/s

t=3 s

a=9.8 m/s²

Using 2nd equation of motion [FREE FALL]

H=1/2×a×t²

H=1/2×9.8×9

H=44.1

So the distance covered in 3 sec is 44.1 m

Now using the distance covered in nth second formula we get;-

S_(nth)=a/2(2n-1)

S_(nth)=9.8/2(2n-1)

S_(nth)=4.9(2n-1)

So in the question It is given that distance in first 3 second(44.1 ) = distance in last second[4.9(2n-1)]

A.T.Q,

44.1=4.9(2n-1)

2n-1=44.1/4.9

2n-1=9

2n=10

n=5 s

So the total time is 5 seconds

Answer 2

Answer:

Given Information,

• Initial Velocity (u) = 0 m/s
• Acceleration = g = 10 m/s²
• Displacement in 3 seconds = Displacement in the last second

To Find,

• Total time of free fall by the object.

Solution,

Using the Second Equation of motion we can calculate the displacement in the first 3 seconds. Hence we get,

⇒ s = ut + 0.5 at²

⇒ s = 0(3) + 0.5 × (10) × (3)²

⇒ s = 0 + 5 × 9

⇒ s = 45 m   ...(i)

Hence the displacement covered in the first 3 seconds is 45 m.

Now let us assume the last second to be 't'.

We know that, the formula to calculate displacement traversed in a nth second is given as:

$\boxed{ \bf{s_n = u + \dfrac{1}{2} \times a \times (2n-1)}}$

Substituting n = t and all the given information we get:

⇒ s(t) = 0 + 0.5 × 10 × (2t - 1)

⇒ s(t) = 5 ( 2t - 1 )

⇒ s(t) - 10t - 5   ...(ii)

According to the question, (i) is equal to (ii). Hence we get:

⇒ 45 = 10t - 5

⇒ 45 + 5 = 10t

⇒ 50 = 10t

⇒ t = 50/10

⇒ t = 5 seconds

Hence the total time of fall is 't' which is 5 seconds.