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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 10335 Points
2 years ago
Dear student

The answer is option c

Regards
Vikas TU
14149 Points
2 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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