Question

If an ball is dropped in air and it reaches to ground after 10 second.What is the disrance covered by the ball in 1 sec , 2 sec and 3 sec . Take g =9.8 m/s . Use eq. s=ut+1/2at^2​

Answer 1

Answer:

Explanation:

1) t=√2h/g

1 = 2× h/9.8

h= 4.9 m

2) t=√2h/g

2^2 = 2× h/9.8

h= 4× 9.8/2

h= 2× 9.8

h = 19.6 m

3) t=√2h/g

3^2 = 2× h/9.8

h= 9× 9.8/2

h= 4.5× 9.8

h = 44.1m

Answer 2

Explanation:

Given :

u = 0m/s

g = 9.8m/s

t = 1s , 2s , 3s

To find : distance ( s) = ?

In 1st case , we take time as 1sec :

s = ut + 1/2 at^2

s = 0m/s * 1s + 1/2 * 9.8m/s * ( 1)^2

s = 4.9m

In 2nd case , time as 2sec :

s = ut + 1/2at^2

s = 0m/s * 2s + 1/2 * 9.8m/s * ( 2s)^2

s = 1/2 * 9.8m/s * 4s^2

s = 19.6m

In 3rd case , time as 3sec

s = ut + 1/2at^2

s = 0m/s * 3s + 1/2 * 9.8m/s * ( 3s)^2

s = 1/2 * 9.8m/s * 9s^2

s = 4.9 m/s * 9s^2

s = 44.1m

hope it helps :-