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a ball is droped from a height of 45m.2 secs later,another ball is thrown downward from the same height.if both hit the ground simultaneously,find the difference between velocities of the two balls,just before hitting the ground. take g=10m/s
Case 1:-
For the first ball
Initial velocity u = 0
Height h = 45 m
Acceleration due to gravity g = 10 m/s^2
Time t = ?
h = ut + 1/2 gt^2
45 = 0 + (1/2)*(10t^2)
45 = 5t^2
t^2 = 45/5
= 9
t = root9
    = 3s
final velocity v1, just before hitting the ground = gt [u=0]
= 10*3
= 30m/s.
Case 2:-
The second ball was thrown 2 seconds later, therefore the time taken by the second ball to travel 45 m is 3 - 2 = 1 s.
Initial Velocity of the second ball u = ?
h = ut + 1/2 gt^2
45 = u*1 + (1/2)*(10)
45 = u + 5
u = 45-5
  = 40m/s. 
final velocity v2 = u+gt
  = 40+(10*1)
  = 40+10
  = 50m/s.
Difference between velocities = v2-v1
      = 50-30
      = 20m/s.


 


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