Question

what is the distance covered by freely falling body during first 3 second of its motion G is equal to 10 metre /sec ^2​

Answer 1

Answer:

Explanation:

given:

u= 0

v= ?

s=?

t= 3 secs

a=10m/s^2

by first eqn of motion,

v=u+at

v= 0+10*3

v=0+30

v= 30m/s

by 3rd eqn of motion,

2as= v^2-u^2

= 2*10*s= 30^2- 0

= 20s= 900

= s= 900 divided by 20

= 450m

hope it helps

Answer 2

Answer:

34

Explanation:

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