A particle Q of mass m falls under gravity in a medium whose resistance to motion is of magnitude mkv2 , where v is the speed of Q and k is a positive constant. Given that the maximum speed of Q is U, show that U=Square root of g/k. g is the acceleration due to gravity. Given that Q is projected vertically upwards with speed V(>U), show that the speed of Q is equal to U when the height of Q above the point of projection is (U2/2g)ln[(1/2)(1+ V2/U2)] Find in terms of U, V and g, the time taken for the speed of Q to decrease from V to U.
A particle Q of mass m falls under gravity in a medium whose resistance to motion is of magnitude mkv2 , where v is the speed of Q and k is a positive constant. Given that the maximum speed of Q is U, show that U=Square root of g/k. g is the acceleration due to gravity. Given that Q is projected vertically upwards with speed V(>U), show that the speed of Q is equal to U when the height of Q above the point of projection is (U2/2g)ln[(1/2)(1+ V2/U2)] Find in terms of U, V and g, the time taken for the speed of Q to decrease from V to U.
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A particle Q of mass m falls under gravity in a medium whose resistance to motion is of magnitude mkv2 , where v is the speed of Q and k is a positive constant. Given that the maximum speed of Q is U, show that U=Square root of g/k. g is the acceleration due to gravity.
Given that Q is projected vertically upwards with speed V(>U), show that the speed of Q is equal to U when the height of Q above the point of projection is (U2/2g)ln[(1/2)(1+ V2/U2)]
Find in terms of U, V and g, the time taken for the speed of Q to decrease from V to U.
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