Question

A boy throws n balls per second at regular intervals.when the first Ball reaches the maximum height he throws the second one vertically up . the maximum height reached by each ball is

Answer 1

Answer: $h\quad =\quad 1.225\quad m$

From the given data since second ball thrown when the first ball reaches maximum height, so let us consider 2 balls are thrown for every second.

When one reaches the maximum height therefore to reach the maximum height, time taken be 0.5 sec.  

The equation of motion, v = u + at

At maximum height h the velocity becomes zero,

                        $v\quad =\quad 0\quad \\\Rightarrow \quad 0\quad =\quad u\quad -\quad g\quad \times \quad 0.5\quad \\\Rightarrow \quad u\quad =\quad 0.5\quad g$

Applying other equation of motion,

                         $v^{ 2 }\quad =\quad u^{ 2 }\quad +\quad 2as\quad \\\Rightarrow \quad 0\quad =\quad u^{ 2 }\quad -\quad 2gh\quad \\\Rightarrow \quad u^{ 2 }\quad =\quad 2gh\quad \\\Rightarrow \quad u\quad =\quad \sqrt { 2gh }$

Now we know that u = 0.5 g as well as $u\quad =\quad \sqrt { 2gh }$

On equating them both, we get,

$0.5\quad g\quad =\quad \sqrt { 2gh }$

$0.25\quad { g }^{ 2 }\quad =\quad 2gh$

$0.25\quad g\quad =\quad 2h$

$h\quad =\quad 0.125\quad g$

$h\quad =\quad 1.225\quad m$