Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

CBSE - IX - Physics - Gravitation

Numerical based on gravitation

Asked by raghavendran7 10th March 2010, 3:20 PM
Answered by Expert

Answer:

Given, H = 100 m
Let the stone thrown up be stone 1
v1 = 25 m/s
a1 = - g = - 10 m/s2 (as it is moving upward)

Let the stone released from H be stone 2
v2 = 0
a2 = g = 10 m/s2  (taking g = 10 m/s2 approximately)

When the two stones meet then,

S1+S2 = H

i.e. (v1t + 1/2 a1 t2 ) + (v2 t + 1/2 a2 t2 ) = H

On substituing the values we get,

          (25xt - (1/2)10xt2 ) + (0xt + (1/2)10xt2 ) = 100

On solving we get,

            25xt = 100
i.e.             t = 4 seconds.

Thus the two stones will meet after 4 seconds.

Hope this helps.
TopperLearning.com

Answered by Expert 23rd April 2010, 5:02 PM

Rate this answer

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Report an issue
Your answer has been posted successfully!

Related Question

Answered by Expert
Answered by Expert
Answered by Expert
Answered by Expert
Answered by Expert

Latest Questions

NEET NEET Chemistry

qsnImg
Asked by amitjena226 24th February 2018, 12:46 AM

NEET NEET Chemistry

qsnImg
Asked by amitjena226 24th February 2018, 12:43 AM

ICSE X Mathematics

qsnImg
Asked by sonali 23rd February 2018, 11:41 PM