11. A person throws a ball vertically upwards with velocity 20 m s-1 from the top of a building of height h and simultaneously throws another ball with same velocity in downward direction. If the ratio of the time taken by two balls to reach the ground is 5: 2, then the height of the building is m. (Take g = 10 m s-2) 800 9 (a) (b) 9 800 400 3 (0) (d) 3
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Answered by
2
Answer:
Correct option is
A
Explanation:
Correct option is
A
solution
t=5s
Time to reach maximum height can be obtained from v=u+at
0=20+(−10)t
t=2s
s=ut+0.5at
2
=20(2)+0.5(−10)(2)
2
=20m
Thus, total distance for maximum height is 45 m
s=ut+0.5at
2
45=0+0.5(10)(t
′
)
2
t
′
=3s
Total time= 3+2= 5s
Answered by
0
Answer:
t=5seconds
Explanation:
Time to reach maximum height can be obtained from v=u+at
0=20+(−10)t
t=2s
s=ut+0.5at
2
=20(2)+0.5(−10)(2)
2
=20m
Thus, total distance for maximum height is 45 m
s=ut+0.5at
45=0+0.5(10)(t')^2
t'=3 sec
total time= 3+2= 5 sec
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