A ball thrown vertically upward returns to thrower after 6s. The ball is 5m below the highest point at t = 2s. The time at which the body will be at same position, (take g = 10m/s^2)
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Answer:
4 sec
Explanation:
The easiest method to solve this question is:
Time of ascent = time of descent
T A = T B = total time taken to back to original position.
T A and TB are time when both balls were at same position.
TA= 2sec (given)
Total time = 6sec (given)
2+ TD = 6
TD = 4 sec
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