Question

A 500g stone is thrown up with velocity of 5ms^-1. Find its
i) P.E at maximum hieght.
ii) K.E at its hit the ground.

Answer 1

Given:-

→ Mass of the stone = 500g = 0.5kg

→ Initial velocity of the stone = 5m/s

To find:-

→ P.E. of the stone at maximum height.

→ K.E. of the stone as it hits the ground.

Solution:-

Number (a) :-

Firstly, let's calculate the maximum height reached by the stone by using the 3rd equation of motion :-

=> v² - u² = 2gh

=> 0 - (5)² = 2(-10)h

=> -25 = -20h

=> h = -25/-20

=> h = 1.25m

Now, we know that :-

P.E. = mgh

=> P.E. = 0.5(10)(1.25)

=> P.E. = 6.25 J

Number (b) :-

After reaching the maximum height, the stone will fall downwards [with u' = 0] So, let's calculate the velocity with which the stone hits the ground (v') by using the 3rd equation of motion :-

=> (v')² - (u')² = 2gh

=> (v')² - 0 = 2(10)(1.25)

=> (v')² = 25

=> v' = 5 m/s

Now, we know that :-

K.E. = 1/2mv²

=> K.E. = 1/2(0.5)(5)(5)

=> K.E. = 12.5/2

=> K.E. = 6.25 J

Thus :-

• P.E. at maximum height is 6.25J

• K.E. as it hits the ground is 6.25J .

Answer 2

We have,

Mass of stone = 500 g = 0.5 kg

Initial velocity = 5 m/s

i) PE at maximum height:-

We know, PE = mgh

Also, 2gh = v² - u²

⇒ h = - u²/2g [Final velocity will become 0]

⇒ h = - (5 m/s)²/[2 × (- 10 m/s²)]

[g = - 10 m/s² (Upward)]

⇒ h = (25 m²/s²)/(20 m/s²)

⇒ h = 1.25 m

∴ PE at max height = mgh = (0.5 kg)(10 m/s²)(1.25 m)

⇒ PE = 5 × 1.25 Nm

⇒ PE = 6.25 J ...(i – Answer)

ii) KE at it's hit in the ground:-

We know, KE = ½mv²

Also, 2gh = v² - u²

⇒ v² = 2gh

⇒ v = √2gh = √2 × 10 m/s² × 1.25 m

⇒ v = √25 m²/s² = 5 m/s

∴ KE = ½ × (0.5 kg) × (5 m/s)² = 6.25 Nm

⇒ KE = 6.25 J ...(ii – Answer).