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Question

A particle is projected vertically upwards with velocity 40ms−1. Find the displacement and distance travelled by the particle in 6 s.
[take g=10m/s2]

A
60 m,100 m
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B
60 m,120 m
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C
40 m,100 m
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D
40 m,80 m
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Solution

The correct option is B 60 m,100 m
Initial upward velocity u=40 ms1
Let after time t, the particle reaches the maximum height.
So final velocity v=0
Acceleration a=g
Using v=ugt
0=4010t t=4 s
Thus after 4 s, the particle starts the downward motion for another 2 s
  • Distance traveled in upward motion:
Here we take a=g as it is downward (negative) and distance is upwards (positive)
S1=ut12gt2 where t=4 s
S1=40×41210×42=80 m

  • Distance traveled in downward motion
Here we take a=g as it is downward (positive) and distance is also (positive)
S2=ut+12gt2 where t=2 s and u=0
S2=0+1210×22=20 m

  • Total distance covered in 6 s: ST=80+20=100 m

  • Displacement in 6 s:
S=ut12gt2 where t=6 s
S=40×61210×62=60 m

Option A is the answer.

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