 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
`        A body freely falling from rest has a velocity v after it falls through distance h.The distance it has to fall down further for its velocity to becomes double is:`
2 years ago

```							 For first 2 m,v^2 = 0^2 + 2g*2v^2 = 4g After that let it traverse d m futher more when its v becomes double .i.e. 2v(2v)^2 = v^2 + 2gd2gd = 3v^2d = 3v^2/2gput the value of v^2d = 3*4g/2g = 12/2 = 6 meter. RegardsArun (askIITians forum expert)
```
2 years ago
```							equation → v2-u2=2ascase 1--> free fallu=0v2=2ghcase2---from height h to h1velocity changes from v to 2v(2v)2-v2=2gh1 from the two equations we get relation between h and h1 ash1=3*h
```
2 years ago
```							Let height from which body is falling when v = b and u = 0 be h'1Let height from which body is falling when v became double and u = v be h'22as=v²-u²(using equation 3 of motion= 2×g×h'1=v²= 2gh'1 = v².  @12gh'2=(2v)² - v². ( This is because we have to make v double to find answer)=2gh'2=4v²-v²=2gh'2=3v²=h'2=3v²/2g (put valu of v² from @1)=3×2gh'1/2g = h'2=3h'1=h'2Answer = 3h'1 in which v became double
```
7 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »  ### 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