MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        
A ball is dropped from a high rise platform at t=0 starting from rest . After 6 seconds another ball is thrown downwards from the same platform with a speed v . The two balls meet at t =18 s . What is the value of v.
3 years ago

Answers : (4)

Hasan Naqvi
97 Points
							
Distance between the two balls after 6 secs = ½ * 10 * 62 = 180
Relative velocity of second ball(wrt first ball) so that it can cover this distance = 180/ 12 = 15m/s
 
Velocity of first ball (at t= 6) =  10* 6 = 60
 
v= 60 + 15 =75m/s
3 years ago
Soumyadeep
20 Points
							
How come it's 180/12 
3 years ago
Nikita Agarwal
13 Points
							The distance travelled by 1st ball in 18 sec is h=1/2gt^2=1/2*10*18^2=1620mNow to meet 2nd ball has to cover the same distance in (18-6 ) sec =12secSo for 2nd ballh=vt+1/2gt^21620=v*12+1/2*10*12^2=75m/s
						
one year ago
khushi jangir
14 Points
							diatance travelled by 1st ball in 18 sec=ut+1/2at^2s=0+1/2×10×18×18s=90×18s=1620mtime taken by second ball=18-6=12secso,distance travelled by second ball=ut+1/2gt×tso,1620==12×u+1/2×10×12×12u=135-60u=75m
						
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details