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
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change