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
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Answered by
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
Answered by
1
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 :-
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