Question

A 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

Answer:

• 19.6 m/s is the required final velocity .

Step-by-step explanation:

According to the Question

It is given that,

• Height ,h = 19.6 m
• Initial velocity ,u = 0m/s (as it is released from top of a tower)

we have to calculate the final velocity just before touching the ground .

As we know the Acceleration due to gravity is 9.8m/s²

Now, using 3rd equation of motion

• v² = u² + 2gh

by putting the value we get

⇢ v² = 0² + 2×9.8 × 19.6

⇢ v² = 0 + 19.6 × 19.6

⇢ v² = 384.16

⇢ v = √384.16

⇢ v = 19.6 m/s

• Hence, the final velocity before touching the ground is 19.6m/s.

Answer 2

Answer:

Given :-

• A stone is released from the top of a tower of height 19.6 m.

To Find :-

• What is the final velocity just before touching the ground.

Formula Used :-

$\clubsuit$ 3rd Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2gh}}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• h = Height

Solution :-

Given :

• Height (h) = 19.6 m
• Initial Velocity (u) = 0 m/s
• Acceleration due to gravity (g) = 9.8 m/s

According to the question by using the formula we get,

$\implies \bf v^2 =\: u^2 + 2gh$

$\implies \sf v^2 =\: (0)^2 + 2(9.8)(19.6)$

$\implies \sf v^2 =\: (0 \times 0) + 2 \times 9.8 \times 19.6$

$\implies \sf v^2 =\: 0 + 19.6 \times 19.6$

$\implies \sf v^2 =\: 0 + 384.16$

$\implies \sf v^2 =\: 384.16$

$\implies \sf v =\: \sqrt{384.16}$

$\implies \sf v =\: \sqrt{\underline{19.6 \times 19.6}}$

$\implies \sf\bold{\red{v =\: 19.6\: m/s}}$

$\therefore$ The final velocity just before touching the ground is 19.6 m/s .

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EXTRA INFORMATION :-

❒ 1st Equation Of Motion Formula :-

$\leadsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

❒ 2nd Equation Of Motion Formula :

$\leadsto \sf\boxed{\bold{\red{s =\: ut + \dfrac{1}{2}at^2}}}$

❒ 3rd Equation Of Motion Formula :

$\leadsto \sf\boxed{\bold{\purple{v^2 =\: u^2 + 2as}}}$

where,

• s = Distance Covered or Displacement
• u = Initial Velocity
• v = Final Velocity
• a = Acceleration
• t = Time Taken

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