Question

A rubber ball is shot straight up from the ground with speed v 0. simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

a. at what height above the ground do the balls collide? your answer will be an algebraic expression in terms of h, v 0 and g.

Answer 1

refer your book at least twice or thrice

Answer 2

Given:  

a) Solution:  

Let the balls collide at a height h1 from the ground.  

When the ball dropped from the height h, Then  

$h-h_1=\frac { 1 }{ 2 } g{ t }^{ 2 }.\longrightarrow (eqn 1)$

For the second ball  

$h_1\quad =\quad V_ot-\frac { 1 }{ 2 } g{ t }^{ 2 }\longrightarrow (eqn 2)$

By adding equation 1 and 2 we get,

$h = V_ot$

By substituting the value of h in equation 2  

$h_1\quad =\quad h-\frac { 1 }{ 2 } g(\frac { h }{ { v }_{ 0 } } )^{ 2 }$

b) When ' h' to be maximum then h1 must be zero. .

$h-\frac { 1 }{ 2 } g(\frac { h }{ { v }_{ o } } )2\quad =\quad 0$

By Simplifying the above equation we get,  

${ h }_{ (maximum) }\quad =\frac { \quad { v }_{ o }^{ 2 } }{ (2g) }$

c) Let us consider the maximum height reached by the first ball is

           $h1 }_{ (max)\quad  }=\quad \frac { { v }_{ 0 }^{ 2 } }{ 2g }$

The maximum time required to reach that height is $(\frac { { v }_{ o } }{ g } )$

In this mean time the distance of fall for the second one is  

            $S\quad =\quad \frac { 1 }{ 2 } g{ t }^{ 2 }\quad =\quad \frac { 1 }{ 2 } g\times (\frac { { v }_{ o } }{ g } )^{ 2 }=\frac { { v }_{ o }^{ 2 } }{ 2g }$

            $h_1+S\quad =\quad h\quad =\quad 2[\frac { { v }_{ o }^{ 2 } }{ 2g } ]=\frac { { v }_{ o }^{ 2 } }{ 2g }$