Question

11. A ball is gently dropped from a height of 80 m. If its velocity increases uniformly at the rate of 10 m/s^2, with what velocity will it strike the ground?​

Answer 1

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

$\sf{Given}\begin{cases}\sf{s=-80m}\\ \sf{u=0ms^{-1}}\\ \sf{a=-10ms^{-2}}\\ \sf{v=??}\end{cases}$

$\textbf{\underline{Using formula,,}}</p><p>$

$\mathtt{\boxed{\blue{v^{2}-u^{2}=2as}}}$

$\sf{\implies{v^{2}-(0)^{2}=2\times{-10}\times{-80}}}$

$\sf{\implies{v^{2}=1600}}$

$\sf{\implies{v=\pm\sqrt{1600}}}$

$\sf{\implies{v=\pm{40}}}$

Since,the velocity is downward , therefore it would be negative.

Hence,,

Velocity,,

$\mathtt{\boxed{\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>v</strong><strong>=</strong><strong>-40ms^</strong><strong>{</strong><strong>-</strong><strong>1</strong><strong>}</strong><strong>}}}$

Hope it helps ❣️❣️❣️