A ball is gently dropped from a height of 40 m if its velocity increases uniformly at the rate of 10 M per second square what is the velocity will it strike the ground
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Answered by
1
20√2
u=0
a=10m/s2
s=40m
v2=u2+2as
u=0
a=10m/s2
s=40m
v2=u2+2as
Answered by
1
v = sqrt(u^2 + 2aS)
= sqrt(0 + 2*10*40)
= 20√2 m/s
Velocity when it strikes the ground is 20√2 m/s
= sqrt(0 + 2*10*40)
= 20√2 m/s
Velocity when it strikes the ground is 20√2 m/s
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