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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12th pass
A stone with weight w is thrown vertically upward into air from ground level with initial speed v0..If a constant force f due to air drag acts on the stone throughout its flight ,the speed of stone just before impact with the ground is what?
one year ago

Answers : (1)

24480 Points
a)initial energy = 1/2 m v^2 
work done by air drag is f h 
potential energy at maximum height h is mgh ( or the work done by/against gravirty) 
1/2mv^2 = fh + mgh 
h = mv^2 /2(f+mg) or writing interms of W = mg , replace m on top with W/g and mg in the debominator with W you will get the result 

b) initial kinetic energy = 1/2mv^2 
final energy = 1/2mV2^2 where v2 is the final velocity 
work done by f is 2fh 
1/2mv^2 = 1/2mV2^2 + 2fh since there is no change in potential energy 
1/2 mv2^2 = 1/2mv^2 -2fh 
W v2^2 = Wv^2 -4fgh 
v2^2 = v^2 - 4gf/W * h 
but we know h = v^2/(2g(1+(f/W))) = Wv^2/(2g(f+W) 
v2^2 = v^2 - 4gf/W * Wv^2/(2g(f+W)) 
v2^2 = v^2 - 2f * v^2/((f+W)) = v^2 [(f+W -2f)/(W+f)] = v^2 (W-f)/(W+f) 
v2 = v^2 √((W-f)/(W+f)) 
one year 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

  • 728 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
To Win!!! Click Here for details