Question

11. A particle is projected vertically upward and moves
freely under gravity. If the distance travelled by the
particle in the 4th second of its motion is twice of
that in the 5th second, then the speed of projection
is (g = 10 m/s?]
(1) 60 m/s
(2) 70 m/s
(3) 40 m/s
(4) 55 m/s


Pls help with explanation​

Answer 1

Answer:

the speed of projection = 55 m/s

Explanation:

Using S = ut  + (1/2)at²

Distance Covered in 3 Seconds = 3u + (1/2)(-g)*3²

= 3u - 4.5g

Distance Covered in 4 Seconds = 4u + (1/2)(-g)*4²

= 4u - 8g

Distance Covered in 6 Seconds = 5u + (1/2)(-g)*5²

= 5u - 12.5g

Distance Covered in 4th sec =  Distance Covered in 4 Seconds - Distance Covered in 3 Seconds

=> Distance Covered in 4th sec = 4u - 8g - (3u - 4.5g)  = u - 3.5g

Distance Covered in 5th sec =  Distance Covered in 5 Seconds - Distance Covered in 5 Seconds

=> Distance Covered in 5th sec = 5u - 12.5g - (4u - 8g)  = u - 4.5g

u - 3.5g = 2(  u - 4.5g)

=> u = 5.5g

taking g = 10

=> u = 55 m/s

the speed of projection = 55 m/s