Question

A stone is released from the top of a tower of height 19•6m calculate its final velocity just before touching the ground ....... please help me to solve this sum ​

Answer 1

the final velocity of the stone can be calculated by using formula

v^2=u^2+2as

v^2=0+2×9.8×19.6

v=√19.6×19.6

v=19.6

therefore the final velocity is 19.6m/s

Answer 2

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From the Question,

• Height of the Tower,h = 19.6 m

Since,the stone starts from rest

• Initial Velocity,u = 0 m/s

• Acceleration due to gravity,g = 9.8 m/s² [According to Sign Convention]

To find

Final Velocity of the stone

Using the relation,

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

Putting the values,we get:

$\sf{v {}^{2} - 0 {}^{2} = 2(19.6)(9.8) } \\ \\ \rightarrow \: \sf{v {}^{2} = 19.6 {}^{2} } \\ \\ \rightarrow \: \sf{v = \sqrt{19.6 {}^{2} } } \\ \\ \huge{ \rightarrow \underline{\boxed{ \tt{v = 19.6 \: ms {}^{ - 1} }}}}$

Thus,the final velocity of the stone would be 19.6 m/s