Question

A ball is allowed to free fall from height h. Find the velocity of ball while it will strike the ground. Or at what velocity it will strike the ground.​

Answer 1

$\sf{\large{\underline{\underline{EXPLANATION:}}}}$

★Given,

• Disp.=height=s=-h

★To find,

• Velocity of the ball while it will strike the ground.

★Now,,

• Here,acc. due to gravity,a=-g=-9.8m/s²

• Initial velocity of the ball,u=0m/s

★Using 3rd equation of motion,,

$\mathtt{\boxed{</strong><strong>v^</strong><strong>{</strong><strong>2</strong><strong>}</strong><strong>-u^</strong><strong>{</strong><strong>2</strong><strong>}</strong><strong>=</strong><strong>2</strong><strong>a</strong><strong>s</strong><strong>}}$

★Putting values,,

$\</strong><strong>s</strong><strong>f</strong><strong>{</strong><strong>v</strong><strong>^{2}-</strong><strong>0</strong><strong>^{2}=2</strong><strong>(</strong><strong>-</strong><strong>9</strong><strong>.</strong><strong>8</strong><strong>)</strong><strong>(</strong><strong>-</strong><strong>h</strong><strong>)</strong><strong>}$

$\sf</strong><strong>{</strong><strong>\</strong><strong>i</strong><strong>m</strong><strong>p</strong><strong>l</strong><strong>i</strong><strong>e</strong><strong>s</strong><strong>{v^{2}=</strong><strong>1</strong><strong>9</strong><strong>.</strong><strong>6</strong><strong>h</strong><strong>}</strong><strong>}$

$\sf{\implies{v=</strong><strong>\</strong><strong>s</strong><strong>q</strong><strong>r</strong><strong>t</strong><strong>{</strong><strong>19.6h}}</strong><strong>}</strong><strong>$

$\sf{\implies{v=</strong><strong>\</strong><strong>p</strong><strong>m</strong><strong>\sqrt{19.6h}}}$

★Therefore,,

V=√19.6h

$\rule{200}2$

Hope it helps ❣️❣️