Question

A man throws balls with the same speed vertically upwards one after the other at an interval of $2s$. What should be the speed of the throw so that more than two balls are in the sly at any time? (Given $g = 9.8\, m/s^2$)

• A. Any speed less than 19.6 m/s
• B. Only with speed 19.6 m/s
• C. More than 19.6 m/s
• D. At least 9.8 m/s

Answer 1

The Correct Option is C

The correct option is(C): More than 19.6 m/s

Time taken by ball to reach maximum height 
\(v = u - gT\) 
at maximum height, final speed is zero ie, \(v = 0\) 
so, \(u=gT\) 
or \(T=u/g\) 
In \(2s,\, u-2\times 9.8-19.6\, ms^{-1}\) 
If man throws the ball with velocity of \(19.6\, ms^{-1}\) then after \(2 s\) it will reach the maximum height. When he throws \(2^{nd}\) ball, \(1^{st}\) is at top. When he throws third ball, \(1^{st}\) will come to ground and \(2^{nd}\) will at the top. Therefore, only \(2\) balls are in air. If he wants to keep more than \(2\) balls in air he should throw the ball with a speed greater than \(19.6\, ms^{-1}\)

Answer 2

The height attained by balls in 2-sec is= \(\frac{1}{2}\) x 9.8 x 4 = 19.6mthe same distance will be covered in 2 seconds for a descent Time interval of throwing balls remains the same. So for two balls remaining in the air, the time of ascent or descent must be greater than 2 seconds. Hence the speed of balls must be greater than 19.6 m/sec.

v = u - gt

 

∴ t = \(\frac{u}{g}\)

∴ T = \(\frac{2u}{g}\) ………(1)

∴\(\frac{2u}{g}\) > 4

∴2u > 4g

∴ u > 19.6 m/s

Thus, the Correct answer is option(C).