Question

a body fall 16 meter in first second of motion 48 meter in the second 80 meter in the third and 112meter in the fourth and so on how far does it fall during the 11 second of its motion​

Answer 1

.
Let x be the distance travelled before first observation is made.

u=0 m/s

V= ?

acceleration =a m/s²

By third equation of motion:

V²-u²=2as

V²-0=2ax

x=V²/2a-----(1)

Distance travelled when t= 1:

S=ut+ 1/2at²

20=Vx1+1/2a(1)²

40=2V+a------(2)

V1=V+at

=V+ax1

V1=V+a-----(3)

Distance travelled in next second:

S=ut+ 1/2at²

40=V1x1+1/2a(1)²

80=2V+2a+a

80= 2V+3a ---(equ 4)

By solving equation 2 and 4

40=2V+a

80= 2V+3a

- - -

-----------------

-40=-2a

a=20m/s²

2v =40-a

2v=40-20

2v=20

v=10m/s

now,X=V²/2a

=10*10/2*20

=10/4

=2.5m

∴ Distance travelled before the first observation was taken is 2.5m

mark as brainlest plzzz