A ball is gently dropped from height of 20m. If its velocity in areas is wniformly distributed at rate of 10ms^-2. What velocity it reaches ground?
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Given that,
Distance(s)="20m"
Initial velocity(u)=0m/s
Accelaration="10m/s-2"
Using the relation Distance-Time
s=v2 - u2=2a
s= v2-0=2*10
20=v2-0=20
v2-0=20*20
v2=400(underoot)
v=20 m/s
time=v=u=at
20=0+10
20 upon 10=0+10(t)=
time= 2 secs
Read more on Brainly.in - https://brainly.in/question/114353#readmore
Distance(s)="20m"
Initial velocity(u)=0m/s
Accelaration="10m/s-2"
Using the relation Distance-Time
s=v2 - u2=2a
s= v2-0=2*10
20=v2-0=20
v2-0=20*20
v2=400(underoot)
v=20 m/s
time=v=u=at
20=0+10
20 upon 10=0+10(t)=
time= 2 secs
Read more on Brainly.in - https://brainly.in/question/114353#readmore
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