1. A body falls freely from rest for 6 seconds. Find the distance travelled in the last two seconds. Take g=9.8 m/s
  2. A body is projected vertically upwards with a velocity of 20 m/s. Find the distance travelled by it in 3 seconds. taking g=10m/s

1.

Here,

t = 6 s

u = 0

Distance traversed in 6 s

s1 = ut +  ½gt2

  = ½ × 9.8 × 62

  = 176.4 m

Let t1  be the time elapsed before last two seconds

t1 = 6 – 2 = 4 s

Distance traversed in 4 s

 s2 = ut +  ½gt2

  = ½ × 9.8 × 42

  = 78.4 m

Distance covered in last two seconds

= s1 – s2

= 176.4 – 78.4

= 98 m

   

2.

Here,

u = 20 ms-1

g = 10 ms-2

t = 3 s

s = ut + ½ (-g)t2

=> s = 20×3 – ½×10×32

=> s = 60 – 45

=> s = 15 m

Distance traversed = 15 m

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1.

s = 0.5 g t 2

Distance travelled in first 4 seconds  = 0.5 * 10 * 42 = 80 m 

Distance travelled in 6 seconds  = 0.5 * 10 * 6= 180 m 

Distance travelled in last 2 seconds = 180 - 80 = 100m

2.

s = ut - 0.5 g t2

= 20*3  - 0.5*10*32

= 60 - 45

= 15 m

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