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: 11
        
a ball thrown up is caught after 6s.the height to which the ball has risen is
7 months ago

Answers : (3)

Bibhu prashad nayak
21 Points
							
Hello friend..!!!
in the above question , we should calculate the height attained by the ball ,
we know H = ut - 1/2gt²
u = ?? 
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 gt²
H = (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 m

thats it.......
7 months ago
AKSHANSH PAREEK
68 Points
							
accleration of body  will be g=10m/sec2
total time of flight=6sec
height to which ball has risen = total distance/ 2
final velocity=0 as body comes to rest
now v=u+gt
0=-u+10*6
u=60m/sec
now v2=-u2+2gs
s=90m
height to which  ball has risen=s/2
=45m
 
 
 
7 months 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
6 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

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