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A projectile is thrown with a velocity of 10sqrt(2) ms^(-1)at an angle of 45° with the horizontal. The time interval between the moments when the speeds are sqrt(125) ms^(-1) is (g=10ms^(-2)) |
Answer» SOLUTION : `u_(n) = 10, u_(y)=0` `v^(2) =v_(x)^(2) + v_(y)^(2)` `125 = 100 +v_(y)^(2)` `Deltat =(2v_(y))/g =(2 xx 5)/10 = 1s` |
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