

InterviewSolution
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A ball thrown vertically upwards with a speed of 19.6 ms-1 from the top of a tower returns to earth in 6 sCalculate the height of the tower |
Answer» u = 19.5 m/s t = 5 sec g = 9.8 m/s² Total distance covered while going up = S total distance covered while coming down = h+S At topmost point v = 0 v = u - gt 0 = 19.5 - 10 (9.8) t = 2 sec (appx) Now, The velocity of ball when it falls down a distance S after reaching topmost height (h+S) will be same as initial velocity (calculate yourself), and the time taken will again be 2 sec. So, total time elapsed = time for going up S + time for coming down S = 2+2 = 4 sec Total time = 5 sec Remaining time = 5-4 = 1 sec here u = 19.5 m/s(downwards) , t = 1 sec h = ut + 1/2 g t² h = 19.5 (1) + 1/2 (9.8) (1) = 19.5 + 4.9 = 24.4 m thankyou |
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