Question

a person can throw a particle to maximum height of hm vertically then the maximum distance up to which he can throw the partial horizontally.

need step by step answer ​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{2h }}}$

$\rule{200}{4}$

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

Let consider person throw a stone vertically with initial velocity u.

The maximum height attained by the stone is h.

At this height the velocity of the stone is zero, i.e  

$\bold{v = 0 }$

Using Newton 2nd Law of motion

$\bf\bold{v^2 - u^2 =2gh}$

Here g is negative due to downward acceleration

$\bf\bold{\therefore\:v^2 - u^2 =2(-g) h}$

$\bold{\Rightarrow \:0 - u^2 =-2gh}$

$\bf\bold{\Rightarrow h = \frac{u^2}{2g}}$ - - - - - - (1)

We know that

Maximum horizontal range, $\bf\bold{R_{max} = \frac{u^2}{g}}$

By using (1)

$\bf\bold{\therefore R_{max} = 2h}$

Hence, the max distance along horizontal is 2h