A man throws a ball in air in such a way that when the ball is in its maximum height he throws another ball. If the balls are thrown after the time difference of 1 sec, then what wilt be the height attained by them
A. 19.6 m B. 9.8 m C. 4.9 m D. 2.45 m
Manvendra Singh chahar , 11 Years ago
Grade Upto college level
2 Answers
Saurabh Koranglekar
Last Activity: 5 Years ago
Dear student
The answer is option c
Regards
Vikas TU
Last Activity: 5 Years ago
It is given that he throws the next ball after the previous one reaches to top.
It is also given that he is throwing 2 balls each second.
Which means after 0.5 second, the ball reaches its maximum height.
Now,
h=?
u=v+gt (v=u-gt)
u=0+10*0.5
u=5 m/s
v=0 m/s
Putting these in the equation v^2 - u^2=-2gh
0–5^2 = -2×10×h
h=1.25 m.
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