(f) time to drop off is ,t = 3 s
We know that ,
Distance coverd by an object dropped from hight is , H = ut + (1/2)gt^2
(H = height covered , u= initial velocity, t= time,
g= acceleration due to gravity =9.82 m/s^2)
Now , H = 0× t + (1/2)×9.81×3^2
= 0+ 44.145 = 44.145 m
So, the height of bridge is, H = 44.145 m
4.(a) initial velocity ,u = 20 m/s
Time taken , t= 4 s, acceleration, a= 4 m/s^2
Now, distance covered,
D = ut + (1/2)at^2
Or, D = 20×4 + (1/2)× 4×4^2
= 80 + 32 = 112 m
4.(b) Effects of force on an objects
1. To change the shape of objects
2. To move or stop the objects
3. To accelerate or deccelerate the objects
4.(c) Demerits of friction
1. Oppose the motion
2. Loss of energy as heat
Way for reducing friction are
1.POLISHING
2.LUBRICATION
(f) In order to find the height of a bridge, a rock was dropped off the...
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