badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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: 8

                        

a ball of mass m kg is thrown vertically upwards. Another ball of mass 2m is thrown at angle θ with the vertical. Both stays in air for the same period . the heights reached by them are in the ratio of (a) 2:1 (b) 1:! © 1:cos θ (d) 1:sinθ

2 years ago

Answers : (1)

Arun
24741 Points
							
height obtained by an object thrown up with u velocity (h) = u^2/2g
 
Hence, ratio of the heights (h1/h2) = u1^2/u2^2
Also, since they remain in the air for same duration. 
So, using first law of motion v = u + at
v = 0, t = -u/a, Now here, since the a is same as equal to g and t is same, so u needs to be same. 
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

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