Question

A body of mass 5 kg is thrown vertically upwards with a speed of 10 m/s. What is its kinetic energy when it is thrown? Find its potential energy when it reaches at the highest point. Also find the maximum height attained by the body.

(g= 10 m/s²)

Plzzz help!❤​

Answer 1

Answer:

KE=mv ^2

= 1/2 x5x10x10

= 250

PE=mgh

Answer 2

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⛄ Answers:-

☘ Kinetic energy of the body when thrown up= 250 J.

☘ Potential energy of the body at the highest point= 250J.

☘ Maximum height attained by the body = 5m.

⛄ Explanation:-

☘ We have,

★ M = 5 kg,

★ v = 10 m/s.

Now, we know that,

★ K.E. = 1/2 mv²

= 1/2 × 5 × (10)²

= 250 J.

Now, when the body reaches at the highest point, its kinetic energy (K.E.) becomes zero and its entire K.E. is converted into P.E. [But the value remains the same..(In accordance with the Law of Conservation of energy as proposed by Sir Einstein)]

Thus,

→ P.E. of the body

= 250 J.

We know that,

★ P.E. = mgh.

Now, on substituting the known values in the given formula, we get,

250 J = 5 × 10 × h

and that gives,

☞ h = 5 m.

Therefore, the maximum height attained by the body is 5 m.

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