A ball is dropped from the roof of a tower of height h.The total distance covered by it in the last second of its motion is equal to the distance covered by it in the first 3 seconds.What is the value of h?
Asked by Aarthi Vishwanathan | 8th May, 2012, 09:53: PM
Expert Answer:
Distance in first 3 second = 45m
Let t be time after dropping ball, V be speed of ball
Through calculation by velocity graph, distance travelled by ball between 4s and 5s = 45m
hence, last second =5s.
h = (5)(50)(1/2)
= 125m
Let t be time after dropping ball, V be speed of ball
Through calculation by velocity graph, distance travelled by ball between 4s and 5s = 45m
hence, last second =5s.
h = (5)(50)(1/2)
= 125m
Answered by | 9th May, 2012, 10:17: AM
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