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A particle is projected with a velocity of sqrt(2) m//s at an angle of 45^(@) with the horizontal. Find the interval between the moments when speed issqrt(125)m//s |
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Answer» Solution :`(g=10 m//s^(2)), u_(n)=10 , u_(y)=0` `V^(2)=v_(X)^(2)+v_(y)^(2)` `125=100 +v_(y)^(2)` `v_(y)=5 Delta =(2v_(y))/(g)=(2xx5)/(10)=1s` 
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