×

#### Thank you for registering.

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

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
```
The height from which a body is released numerically equal to the velocity aquired finally then that height is equal to
The height from which a body is released numerically equal to the velocity aquired finally then that height is equal to

```
3 years ago

Rohan
23 Points
```							As s= ut + a(t)(t)/2 but in this case u=0 as the body is dropped simply.Also v=u +at.Putting u=0 in both the cases we get, s=a(t)(t)/2   and v=at.As given s=v ,and cancelling a&t from both sides we get t=2 sec.Now a=g ,t=2 hence s=9.8× 4/2=19.6m.Hence height= 19.6 m.
```
3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### 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