What will be the velocity of a 0.23 kg arrow having a potential energy of 61 J after it is shot from a bow?

530.43m/s

Explanation:

Given data

Mass m = 0.23kg

PE= 61J

From the expression for potential energy

PE= mgh

after it was shot

PE=KE

KE= 1/2mv^2

So

PE= 1/2mv^2

substitute

61= 1/2*0.23*v^2

61=0.5*0.23*v^2

61=0.115v^2

divide both sides by 0.115

v= 61/0.115

v=530.43 m/s

Hence the velocity is 530.43m/s