Question

1 ball is dropped from the height of 80 meters, the distance covered in the fourth second​

Answer 1

s=ut +1/2at^2

80=0.5*9.8*t^2

t^2=80/4.9

t is approx=16^1/2

t=4 sec

NOW.......

A body covers distance in odd no. ratio when is freely falled (galilio's law)

(1:2:3:4)x=80

10x=80

x=10

xistance in last second =40 metres

Answer 2

Given :

Height from where ball is dropped , h = 80 m .

Initial velocity of ball , u = 0 m/s .

To find :

The distance covered in the fourth second​ .

Solution :

We know , by equation of motion :

$s=ut+\dfrac{at^2}{2}$

Here , a = g which is acceleration due to gravity and equal to $g=9.8\ m/s^2$ .

So , putting value of t = 4 s and u = 0 m/s , in above equation .

We get :

$s_4=0\times 4+\dfrac{9.8\times 4^2}{2}\\\\s_4=78.4\ m$

Now , putting t = 3 s .

We get :

$s_3=0+\dfrac{9.8\times 3^2}{2}\\\\s_3=44.1\ m$

So , distance travelled in fourth second​ is :

$D=s_4-s_3\\D=74.8-44.1\ m\\D=30.7\ m$

Therefore , distance covered in the fourth second​ is 30.7 m .

Learn More :

Kinematics

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