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A ball is projected vertically upwards at t = 0. The ball is at same height at time 4 s and 8 s. The maximum height attained by the ball is(g = 10 m/s2)(1) 120 m(2) 80 m(3) 180 m(4) 100 m |
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Answer» answer : option (3) explanation : A ball is projected vertically upwards at t = 0. The ball is at same height at time 4 s and 8 s. The trajectory of the ball obeys the equation , S = ut + 1/2 (-g)t² , [where u is initial velocity of ball,and g is acceleration DUE to gravity. ] so, at t = 4s and 8s , height of ball remains same. putting value of t = 4s S = u(4) + 1/2(-g) × (4)² = 4u - 8g ....(1) putting value of t = 8s S = u(8) + 1/2(-g) × (8)² = 8U - 32g .....(2) from EQUATIONS (1) and (2), S - S = (8u - 32g) - (4u - 8g) or, 0 = (8u - 4u) - (32g - 8g) or, 0 = 4u - 24g or, u = 6G.....(1) at maximum height velocity of ball becomes zero. so, final velocity of ball , v = 0 now using formula, v² = u² + 2(-g)h here, v = 0, u = 6g so, 0 = (6g)² - 2gh or, h = 36g²/2g = 18g when g = 10m/s² then, h = 180m hence, maximum height attained by the ball is 180m |
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