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A stone is thrown with vertical direction with a velocity of 10 m/s. if the acceleration of the stone during its motion is 10 m/s ² in the downward direction what will be the height attained by stone and how much time will it take to recall there. |
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Answer» -INITIAL Velocity of stone = u = 10 ms-¹Acceleration of stone = a = -10 ms-²Let the maximum height attained by body = HHere, Negative SIGN indicates acceleration in downward direction.Using third equation of Motion,v² = u² + 2aH0² = (10)² + 2(-10)H-100 = -20HH = 5 m Again,H = ut + 1/2 at²5 = 10T + 1/2 (-10)t²5t² - 10t + 5 = 05t² -5t - 5t + 5 = 05t(t - 1) - 5(t - 1) = 0(5t - 5)(t - 1) = 0As we know that when product of two TERMS is equal to zero then either first term is equal to zero or second term is equal to zero. 5t - 5 = 0t = 1 sor,t - 1 = 0t = 1 sHence,Maximum height attained by body = H = 5 m Total time for body to come back = 2t = 2s |
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