A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.
A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.
This is a short answer type question as classified in NCERT Exemplar
When the ball dropped from the building u1=0, u2=40m/s
Velocity of the dropped ball after time t
V1=u1+gt
V1 = gt
For ball thrown up u2=40m/s
Velocity of the ball after time t
V2=u2-gt
=40-gt
Relative velocity =v1-v2
=gt- [-40-gt]=40m/s
Similar Questions for you
Please find the solution below:
[h] = ML2T-1
[E] = ML2T-2
[V] = ML2T-2C-1
[P] = MLT-1
According to question, we can write
10 =
Average speed
(d) Initial velocity
Final velocity
Change in velocity
Momentum gain is along
Force experienced is along
Force experienced is in North-East direction.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...

Physics NCERT Exemplar Solutions Class 11th Chapter Three 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering


