Question

a stone of mass 0.2 kg is projected vertically upward with a velocity of 25 m/s . Calculate the maximum height by it above the ground.At what height will it lose 20% of its initial kinetic energy?​

Answer 1

Given :

• Mass of stone = 0.2 kg
• Initial velocity of stone = 25 m/s

To find :

• Maximum height reached
• At what height it will lose 20% of it's initial kinetic energy

Solution :

At first we have ;

• Initial velocity of body (u) = 25 m/s
• Acceleration due to gravity (g) = 10 m/s²

Maximum height reached is given by :

⇔ h = u²/2g

⇒ h = (25)²/(2 × 10)

⇒ h = 625/20

⇒ h = 31.25 m

∴ Maximum height reached = 31.25 m

Now at a particular height 20% loss of kinetic energy = Potential energy of stone.

So atq :

⇒ 20% of (1/2 mv²) = mgh

⇒ 0.2 × 1/2 × 0.2 × (25)² = 0.2 × 10 × h

⇒ (0.04 × 625)/2 = 2h

⇒ 4h = 25

⇒ h = 25/4

⇒ h = 6.25 m

∴ At a height of 6.25 m it will lose 20% of it's initial kinetic energy.

Answer 2

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