A man throws balls with the same speed vertically upwards one after the other at an interval of 2 second. What should be the speed of the throw so that more than two balls are in the sky at any time? (given g=9.8m/s2)

Question

Answer

Answer :- more than 19.6 m/s Let the required speed of through be u m/s. The time taken to reach maximum height t=u/g ; For two balls to remain in air at any time, t must be greater than 2. ; ∴ u/g >2 ⇒ u > 2(g) ; u > 19.6 m/s

Answer

The man is throwing the stones one by one after two seconds interval so if more than two stones are in the air so we consider that two atones are in the air so the condition comes one is about to reach at ground.second is in maximum height and the third one os about to throw by men so two are in the air at t=4sec  when we calculate the velosity of the second stone v=u+at where t is time takken by the stone to reach maximuym heightr is 2 sec so v=0=u-g.2so u=2g=19.6m/secsamely  we can take it second way by first stone where time taken is 4 sec so -u=u-g.4 also comes u=2g so the velosity should be greater than 19.6m/sec

Answer

Firstly I would like to appreciate your patience for giving such a long answer

Answer

Introduction
To have more than two balls in the sky at any time, we need to ensure that the time taken by each ball to reach the ground is greater than 2 seconds. This can be achieved by throwing the balls with a certain minimum velocity.

Calculating time taken by each ball
We know that the time taken by a ball to reach the ground is given by the formula:
t = sqrt(2h/g)
where h is the height of the ball from the ground and g is the acceleration due to gravity.

Since the ball is thrown vertically upwards, the initial velocity is zero. Therefore, the height of the ball at any time t is given by:
h = ut - (1/2)gt^2
where u is the initial velocity.

Substituting this value of h in the previous formula, we get:
t = sqrt(2u/g) - sqrt(2(g/2)(2u/g))

Simplifying this expression, we get:
t = sqrt(2u/g) - sqrt(u/g)

Determining minimum velocity
To ensure that the time taken by each ball to reach the ground is greater than 2 seconds, we need to solve the inequality:
sqrt(2u/g) - sqrt(u/g) > 2

Squaring both sides and simplifying, we get:
u > 19.6 m/s

Therefore, the minimum velocity with which the balls should be thrown vertically upwards is 19.6 m/s.

Conclusion
To have more than two balls in the sky at any time, the balls should be thrown with a minimum velocity of 19.6 m/s vertically upwards. This ensures that the time taken by each ball to reach the ground is greater than 2 seconds, and hence, there are more than two balls in the sky at any time.

Answer

no this question is right

Answer

bhai,ques.incomplete aur glt, dono h

Answer

How is v= -u in the last equation? can somebody plz answer?

Similar Class 11 Doubts

two projectiles are thrown from the same point simultaneously with same velocity 10m/s .one goes straight vertically while other at 60 degree with the vertical. what will be the distance of separation between the two after 1 second of their throw

two projectiles are thrown from the same point simultaneously with same velocity 10m/s .one goes straight vertically while other at 60 degree with the vertical. what will be the distance of separation between the two after 1 second of their throw.

With what speed should a body be thrown upwards so that the distance traversed in 5th second and 6th second are equal? correct answer: 49m/s?How to solve?

Top Courses for Class 11View all

Chemistry Class 11

CUET Mock Test Series

English Class 11

Chemistry Practice Tests: CUET Preparation

CUET Mock Test: Humanities Subjects 2025

Top Courses for Class 11

Top Courses for Class 11

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams