A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for (1)3s (2)5s (3)7s (4)9s?

Question

Answer

Free Fall of Body

Problem Statement:

A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for

(1) 3s

(2) 5s

(3) 7s

(4) 9s

Solution:

Concept of Free Fall

Free fall is the motion of an object under the influence of gravity only. When an object falls freely, it accelerates downwards at a constant rate called the acceleration due to gravity (g).

Distance Covered in Free Fall

Distance covered by an object in free fall can be calculated using the formula:

d = 1/2 * g * t^2

where d is the distance covered, g is the acceleration due to gravity and t is the time for which the object falls.

Given Information

The problem statement states that:

The body covers as much distance in the last second of its motion as covered in the first three seconds.

Derivation

Let the time for which the body falls be t seconds.

According to the problem statement, the distance covered in the last second is equal to the distance covered in the first three seconds.

Therefore, the distance covered in the first three seconds can be calculated as:

d1 = 1/2 * g * 3^2 = 4.5g

Let the distance covered in the last second be d2.

Therefore, the distance covered in the last second can be written as:

d2 = 1/2 * g * (t-1)^2 - 1/2 * g * (t-2)^2

Equating d2 to d1, we get:

1/2 * g * (t-1)^2 - 1/2 * g * (t-2)^2 = 4.5g

Simplifying the above equation, we get:

t^2 - 7t + 10 = 0

Solving the above quadratic equation, we get:

t = 5s or t = 2s

Since the body has fallen for some time, t cannot be 2s.

Therefore, the time for which the body falls is 5s.

Answer

Therefore, the correct option is (2) 5s.

Answer

For 1 st 3 sec. of motion S = 1/2 at^2= 9a/2 distance in last sec S=  u+ a/2(2t-1 ) so a/2 (2t-1)= 9a/2  by solving this t = 5 sec

Answer

It's 5s by using equation s=ut +1/2 gt^2

Answer

Please explain

Similar NEET Doubts

A body falls freely from rest it coveres as much distance in the last second of its motion as covered in the first three seconds the body has fallen from a time of

A ball falls freely from rest and the total distance covered by it in the last second of its motion is equal to the distance covered by it in first five seconds of its motion. the total time of the stone when it is in motion would be

A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three second of its motion.the stone remains in the air of?

A body starts from rest with uniform acceleration and travels a distance of 9 is 525 of Total distance in last second of its journey find how much time body is in motion and how much distance travel is if it covers 6 metre in 1 second?

A body drop from the top of a tower covers a distance 7x in last second of its journey,where x is the distance covered in 1st second.how much time it will take to reach the ground?

Top Courses for NEETView all

Biology Class 11

Daily Test for NEET Preparation

NEET Mock Test Series 2025

Topic-wise MCQ Tests for NEET

Physics Class 11

Top Courses for NEET

Top Courses for NEET

Suggested Free Tests

Related Content

About NEET

NEET Preparation Material

How to Prepare for NEET Exam?

NCERT Based Study Material

NEET Mock Tests

Other Popular Exams

Company