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 ball thrown up is caught after 6s.the height to which the ball has risen is`
one year ago

21 Points
```							Hello friend..!!!in the above question , we should calculate the height attained by the ball ,we know H = ut - 1/2gt²u = ?? t = 6 sec h = ??we know that ,v = u + at  ( but here v = -u before reaching the ground )-u = u + at -2u = at  ( a = g = 9.8 and t = 6 )u = ( 9.8 x 6 ) / 2 u = 29 . 4 m / s -----------------------( 1 ) we know,H = ut - 1/2 gt²H = (29.4)(3) - 1/2(9.8)(3x 3 )here i considered t = 3 sec because , the question asked is for max height so if the total time taken for the journey is 6 sec then half of it is 3 sec.H = 44.11 mthats it.......
```
one year ago
AKSHANSH PAREEK
68 Points
```							accleration of body  will be g=10m/sec2total time of flight=6secheight to which ball has risen = total distance/ 2final velocity=0 as body comes to restnow v=u+gt0=-u+10*6u=60m/secnow v2=-u2+2gss=90mheight to which  ball has risen=s/2=45m
```
one year ago
AKSHANSH PAREEK
68 Points
```							answere which i had given is also correct as i have taken g=10m/sec2 not 9.8m/sec2  please approve my answer that was correct
```
one year 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