Question

A 5 kg ball is dropped from a height of 10 m. (a)find initial potential energy of the ball. (b)calculate the velocity before it reaches the ground​

Answer 1

Given :

Mass of ball, m = 5kg

Height, h = 10 m

To find :

a)Initial potential energy of ball.

b) Velocity of ball before reaching the ground.

Solution :

➠ Potential energy, P.E. = mgh

Where, g = gravitational acceleration = 9.8 m/s²

So,

⇒ P.E. = 5 × 9.8 × 10

⇒ P.E. = 50 × 9.8

⇒ P.E. = 490 J

∴ Initial potential energy = 490 J

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Now as ball reached ground, so whole potential energy will convert to kinetic energy.

So atq,

➠ Kinetic energy, K.E. = 490 J

➠ Kinetic energy, K.E. = 1/2 mv²

So atq,

⇒ 490 = 1/2 × 5 × v²

⇒ 490 = 5/2 × v²

⇒ 490 × 2/5 = v²

⇒ 196 = v²

⇒ v = √196

⇒ v = 14 m/s

∴ Velocity of ball before reaching ground = 14 m/s

Answer 2

Given

• Mass of ball (m) =  5 Kg
• Height (h) = 10 m

To find

1. Initial potential energy of the ball
2. Velocity before the ball reaches the ground

Solution

We know potential energy can be find with the formula :-

• PE = mgh

Where :-

• PE - potential enerhy
• m - mass
• g - acceleration due to gravity
• h - height

Substituting the value, we get :-

• PE =  5 × 9.8 × 10
• PE = 490

Hence, the potential energy is 490 J

According to the question, body reached the ground, so potential energy will be converted to kinetic energy.

✮ Energy stored in an object due to its position is Potential Energy. Energy that a moving object has due to its motion is Kinetic Energy.

We know, Kinetic energy =  1/2 mv²

Where :-

• m = mass
• v = velocity

Substituting the values, we get :-

• 490 = 1/2 × 5 × v²
• 490 = 5/2 × v²
• 196 = v²
• v² =  √196
• v = 14

Hence, the velocity is 14 m/s