Published by:
CGP EDU Academic Team
Published on: September 11, 2026
Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms -1 . [use g = 10 ms -2 ] :
Text Solution
Verified by ExpertsThe correct answer is:
D
Let they meet at time t.

= 4 sec
Time taken by ball B to meet A = 2 sec
using S = ut+
at 2

U = 30
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A particle of mass m is dropped from a height h above the ground. At the same time another particle…
The relation between root mean square speed (v rms ) and most probable speed (v p ) for the molar m…
A tuning fork of frequency 340 Hz resonates in the fundamental mode with an air column of length 12…
A force on an object of mass 100g is . The position of that object at t = 2s is after starting fr…
is a vector quantity such that = nonzero constant. Which of the following expressions is true for…
The distance of the Sun from earth is 1.5 x 10 11 m and its angular diameter is (2000) s when obser…