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A particle is projected with a velocity of 10sqrt(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 = 10m//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 t = (2v_(y))/(g) = (2 xx 5)/(10) = 1s`
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