Question

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A body of mass 1 kg drops from the top of towr of height 50m what will be its K.E 10m below the top.

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Answer 1

Given -

• Mass of the body = 1kg
• Height of the tower = 50m

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To find -

• Kinetic energy of the body 10m below the top of the tower.

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Solution -

So firstly, we have to find the velocity of the body at 10m below the top. Here we use the relation given below.

$\: \: \: \: \: \: \: \: \: \:\underline{\bf{\red{v^2 - u^2 = 2gh}}}$

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We have,

• u = 0
• g = 10m/s²
• h = 10m

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Putting the values

$\tt\dashrightarrow{v^2 - (0)^2 = 2 \times 10 \times 10}$

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$\tt\dashrightarrow{v^2 - 0 = 2 \times 100}$

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$\tt\dashrightarrow{v^2 = 200m/s^2}$

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Now for finding the kinetic energy, we use

$\: \: \: \: \: \: \: \: \: \:\underline{\bf{\red{K.E = \dfrac{1}{2} mv^2}}}$

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Substituting values

$\tt\dashrightarrow{K.E = \dfrac{1}{2} \times 1 \times 200}$

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$\tt\dashrightarrow{K.E = \dfrac{1}{2} \times 200}$

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$\tt\dashrightarrow{K.E = 100 J}$

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Hence,

• Kinetic energy of the body 10m below the top is 100 joules.

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Answer 2

A body of mass 1.0kg is dropped from the top of a tower of height 50m. ... Kinetic Energy of the body of 1 kg after a fall through of 10m = ½ m v² = ½ × 1× 40 = 20 Joule.

Hope it helps uh.