Question

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

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Answer 1

this is the answer for this question

Answer 2

Solution :-

According to the equation of motion under gravity

${\large{\bold{{v}^{2}\:-\:{u}^{2}\:=\:2gs}}}$

where :-

• u = Initial velocity

• v = final velocity

• s = height of stone

• g = acceleration due to gravity .

given :-

• u = 0 $\dfrac{m}{s}$

• s = 19.6m

• g = 9.8 $\dfrac{m}{s}^{-2}$

Find :-

• final velocity (v)

Formula used :-

⠀⠀⠀⠀⠀⠀${\boxed{\large{\bold{{v}^{2}\:-\:{u}^{2}\:=\:2gs}}}}$

→ Substituting the values in the above formula , we get

⠀⠀⠀⠀⠀⠀⠀⟹${(v)}^{2}$ - ${(0)}^{2}$ = $2\:×\:9.8\:×\:9.16$

⠀⠀⠀⠀⠀⠀⠀⟹ ${(v)}^{2}$ = 2 × 9.8 × 9.16

⠀⠀⠀⠀⠀⠀⠀⟹ ${(v)}^{2}$ = ${(19.6)}^{2}$

⠀⠀⠀⠀⠀⠀⠀⟹ 19.6 $\dfrac{m}{s}$