Grade 11Mechanics
3rd one sol and procedures for solution detailedly plz quick
3rd one sol and procedures for solution detailedly plz quick
1 Answer
Arun
When it is dropped from 10m,
Initial height = 10m
initial velocity = 0
velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)
Initial height = 10m
initial velocity = 0
velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)
after rebound,
maximum height reached = 2.5m
final velocity at top = 0
initial velocity(just after rebound) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)
maximum height reached = 2.5m
final velocity at top = 0
initial velocity(just after rebound) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)
assuming downward as positive direction
So velocity just before hitting ground = +14.07 m/s
velocity just after hitting ground = -7 m/s
change in velocity = +14.07 - (-7) = 21.07 m/s
time = 0.02s
acceleration = change in velocity/time = 21.07/0.02 = 1053.5 m/s² ~ 1050m/s²
So velocity just before hitting ground = +14.07 m/s
velocity just after hitting ground = -7 m/s
change in velocity = +14.07 - (-7) = 21.07 m/s
time = 0.02s
acceleration = change in velocity/time = 21.07/0.02 = 1053.5 m/s² ~ 1050m/s²
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