Question

A stone is released from the top of a tower of height 50 m. Calculate its final velocity just
before touching the ground.​

Answer 1

Answer:

50*9.8=49 m/s

Step-by-step explanation:

Answer 2

$\large\bold{\underline{\underline{{Given:-}}}}$

${\bullet\:}$ A stone is released from the top of a tower of height 50 m.

$\large\bold{\underline{\underline{{To\:Find:-}}}}$

${\bullet\:}$ Its final velocity just  before touching the ground.

$\large\bold{\underline{\underline{{Solution:-}}}}$

According to the third equation of motion :-

Here,

➣ v = final velocity

➣ u = initial velocity

➣ g = acceleration due to gravity

➣ s = height

$\red{\bigstar}$ Units :-

➣ v = metres per second (m/s)

➣ u = metres per second (m/s)

➣ g = metres per second square (m/s²)

➣ s = metre (m)

$\red{\bigstar}$ According to the question :-

${\bullet\:}$ We should find the velocity (v) just before touching the ground.

Hence,

➣ Initial velocity (u) = 0 m/s

➣ Distance travelled, (s) = 50 m

➣ Acceleration, a = (g) = 10 m/s²

${\bullet\:}$ Let the final velocity be v.

${\bullet\:}$ Substituting the values ........we get ,

$\pink{\longrightarrow \:}$$\sf{v^{2}-u^{2} = 2\times a\times s }$

$\pink{\longrightarrow \:}$$\sf{v^{2}-0^{2} = 2\times 10 \times 50 }$

$\pink{\longrightarrow \:}$$\sf{v^{2}=\sqrt{2 \times 10\times 50 }  }$

$\pink{\longrightarrow \:}$$\sf{v^{2}=\sqrt{20 \times 50 }  }$

$\pink{\longrightarrow \:}$$\sf{v^{2}=\sqrt{1000 }  }$

$\pink{\longrightarrow \:}$$\sf{v=\sqrt{1000 }\:m/s  }$

$\pink{\longrightarrow \:}$$\sf{v=31.62\:m/s}$

Hence,

$\large\boxed{\underline{\purple{\sf \therefore \: Final\:velocity\:just\:before\:touching\:the\:ground=31.62\:  m/s }}}$

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