Question

A stone is released from the top of tower of height 19.6m .Calculate final velocity just before touching the ground......................

Answer 1

Answer:

this is your answer

Explanation:

all are clear

Answer 2

$\boxed{\begin{minipage}{11 cm} Given \\ \\ \ u=0 m/s \\ \\ Height=19.6m \\ \\ g = 9.8m\;s^{-2} \\ \\ To\;Find:- \\ \\ v = ? \end{minipage}}$

$\textbf{\boxed{Third\; Equation\;of\; Gravitation}}$

${\boxed{v^{2} = u^{2} + 2gh}}$

★We can also write it as :-

${\boxed{v^{2} - u^{2} = 2gh}}$

★Substitute the values :-

${\boxed{v^{2} - 0^{2} + 2\times 9.8m s^{-2}\times 19.6}}$

${\boxed{v^{2} = 19.6 ms^{-2}\times 19.6m}}$

${\boxed{v^{2} = (19.6)^{2}m^{2}s^{-2}}}$

${\boxed{v= 19.6ms^{-2}}}$

$\boxed{\begin{minipage}{11 cm} Additional Information \\ \\ \ Final\;Velocity=Initial\; velocity+Acceleration\times Time \\ \\ Momentum = Mass\times Velocity \\ \\ Force=Mass\times Acceleration \\ \\ Force = \dfrac{Change\;in\;Momentum}{Time\;Interval} \\ \\ Impulse = Force\times Time \end{minipage}}$