A particle is dropped from tower. It covers 40 m in last 2 sec .Find height of tower?

Question

Answer

Let the body fall through the height of tower in t seconds. From,     Dn=u+a/2 (2n-1)                                      we have,total distance travelled in last 2 seconds of fall is     D= Dt+D((t-1) ) = [ 0+ g/2 (2t-1)]+ [0+g/2 {2(t-1)-1}] =g/2  ( 2t-1)+g/2 (2t-3)=  g/2(4t-4)                    = 10/2×4(t-1)                                                                         Or, 40 = 20(t-1) or t =2+1 = 3s                                      Distance travelled in t seconds is s=ut +1/2   at^2 =0+1/2×10×3^2=45 m

Similar NEET Doubts

A particle is dropped from a tower .it is found that it travels 45 m in the last second of its journey.find out the height of the tower.

can anyone tell me how to solve the equation formed in this questionquestion is- A ball is dropped from the top of a tower of height h. it covers a distance h/2 in the last second of its motion. how long does the ball remain in air?

A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle?

A particle is dropped from the top of the tower. During its motion it covers 9/25 th part of the tower in the last one sec then velocity with which it will hit the ground?

A body is released from the top of a tower of height H m. After 2 s it is stopped and then instantaneously released. What will be its height after next 2 s ?

Top Courses for NEETView all

Biology Class 11

Daily Test for NEET Preparation

NEET Mock Test Series 2025

Topic-wise MCQ Tests for NEET

DPP: Daily Practice Problems for NEET

Top Courses for NEET

Top Courses for NEET

Related Content

About NEET

NEET Preparation Material

How to Prepare for NEET Exam?

NCERT Based Study Material

NEET Mock Tests

Other Popular Exams

Company