Contact

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

# A ball is droped from a high rise platform t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v . The two ball meet at t=18s . What is the value of v ?      1)74m/s          2)64m/s .  3)84m/s.    4)94m/s

Asked by Kajalmandal64 11th May 2018, 2:53 PM
First ball starting from rest has travelled a distance S1 in 18 seconds.

S1 = (1/2)×9.8×18×18 = 1587.6 m

Second ball thrown with initial velocity V,  has travelled a distance S2 in 12 seconds;

S2 = V×12+(1/2)×9.8×12×12= (12V+705.6) m;

since both balls meet at t=18s, then S1=S2;
hence 12V+705.6 = 1587.6; solving for V, we get V = 73.5 m/s
Answered by Expert 11th May 2018, 3:43 PM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer 2/10

RELATED STUDY RESOURCES :

### Latest Questions

CBSE X Biology
Asked by Sureshgowdayd548 28th October 2020, 10:50 PM
CBSE X Biology
Asked by reenacharan2008 28th October 2020, 9:11 PM