LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
431 views
in Physics by (72 points)
edited by

At t = 0, an elevator departs from the ground with uniform speed. At time T1 a child drops a marble through the floor. The marble falls with uniform acceleration g = 9.8 m/s^2, and hits the ground T2 seconds later. Find the height of the elevator at time T1.

Please log in or register to answer this question.

1 Answer

+1 vote
by (26.4k points)

Let us assume,

The velocity with which the elevatar ascends be u', the height attained by elevatar at time,T1 be h .

Thus,

u' = \(\frac{h}{T_1}\)

Equation of motion,

h = -u'T2\(\frac{1}{2}gT_2^2\)

∵ u' = \(\frac{h}{T_1}\)

h + \(\frac{h}{T_1}\)T2\(\frac{1}{2}gT_2^2\)

h(1 + \(\frac{T_2}{T_1}\)) = \(\frac{1}{2}gT_2^2\) 

h = \(\frac{1}{2}\)\(\frac{gT_2^2}{(1+\frac{T_2}{T_1})}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...