No products in the cart.

A body dropped from top of …

CBSE, JEE, NEET, CUET

CBSE, JEE, NEET, CUET

Question Bank, Mock Tests, Exam Papers

NCERT Solutions, Sample Papers, Notes, Videos

A body dropped from top of tower fall through 40m during the last two seconds of its fall. The height of tower is( g= 10 m/s^2
  • 3 answers

Shanika Sharma 1 year, 11 months ago

Let us suppose that during travelling 40 m in last 2 sec its initial velocity was u. Then, ut+1/2 gt² =40 =2u-19.6=40 u=29.8 m/sec Now considering from top to this position, u=0and v=29.8m/s Then by, v²−u²=2gh we get, 29.8²-0²=2×9.8×h h=45.3m So, total height of tower =45.3+40=85.30 m

Sanjay Gangwar 2 years ago

No

Aryan Singh 2 years ago

Sol or ans.
http://mycbseguide.com/examin8/

Related Questions

2d+2d =
  • 1 answers
1dyne convert to S.I unit
  • 1 answers
Project report
  • 0 answers
√kq,qpower2 R2
  • 0 answers
Ch 1 question no. 14
  • 0 answers

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App