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Grade 12th passMechanics

a ball is dropped freely while another is thrown vertically downward with an initial velocity v from the same point simultaneously.after t second they are separated by a distance of....solution

Srihari

8 Years agoGrade 12th pass

2 Answers

Soumya Ranjan Mohanty

8 Years ago

For the dropped ball, initial velocity is 0,time=t s and acceleration=g ms^-2 downwards. Distance travelled for dropped ball=x1=ut+1/2 gt^2=0+1/2gt^2.

Similarly, for the ball thrown, initial velocity=v ms^-1, time= t s and acceleration=g ms^-2 downwards. Distance travelled for ball thrown with velocity v=x2=ut+1/2 gt^2=vt+1/2 gt^2.

So distance of separation between them=x2-x1=vt+1/2 gt^2-1/2 gt^2=vt,

therefore, distance of separation between the 2 balls is vt unit distance.

Thanks,

Soumya Ranjan Mohanty.

Vikas TU

8 Years ago

Dear Student,

ball1
u=0
v=u+at
=>v=gt

S1=ut+at²/2
=>S1=gt²/2

ball2
u=v
vf=u+at
=>vf=v+gt

S2=vt+gt²/2

distance of separation = S2-S1
=vt.

Cheers!!

Regards,

Vikas (B. Tech. 4^th year
Thapar University)

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