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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

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

Grade:Upto college level

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
Dear student

The answer is option c

Regards
Vikas TU
14149 Points
one year 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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