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A ball dropped from a balloon at rest clears a tower 81m high during the last quater second of its journey. Find the height of the balloon and the velocity of the body when it reaches the ground.WHAT IT MEANS- LAST QUATER SECOND OF ITS JOURNEY. Solution) The last quarter of the jouney menas 1/4th of the total time it requires in its total journey.Let total time be 4t, u = 0.So, v at 3t = 3gt, and at 4t = 4gt.by v² = u² + 2as,16g²t² = 9g²t² + 2g(81)7g²t² = 162g7gt2 = 162t² = 162 / 7gt = 1.5 sec.but time is 4t = 6 sec.So velocity on reaching = 4gt = 58.8 m/sWith this velocity, height of fall = 2t(58.8) = 705.6m
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Q. When the negative feedback is practical to an amplifier of gain 50 the gain falls to 25. Compute the feedback ratio? Af = A / 1 + Aβ 25 = 50 / 1 + 50 β 1 = 2 / 1 + 50
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