Question

a ball of mass 200g falls from a height of 5metres. What
is its Kinetic Energy when it just reaches the ground? (g=9.8m/s^2)​

Answer 1

${\underline{\underline{\huge{\mathfrak{\green{Answer:-}}}}}}$

Hey Mate!!

Given,

m = 200 g

h = 5 m

u = 0 m/s

g = 10 m/s^2

${v}^{2} - {u}^{2} = 2gh \\ = > {v}^{2} - 0 = 2 \times 9.8 \times 5 \\ = > {v}^{2} = 98 \\ = > v = \sqrt{98} \\ = > v = 10 \: (approx.)$

K.E. = 1/2 mv^2

=> 1/2 × 200 × (10)^2

=> 100 × 100

=> 10000 joules

Answer 2

Given :-

• Mass of the ball (m) = 200 g
• Height (h) = 5 m
• Acceleration due to gravity (g) = 9.8 m/s²

To Find :-

• kinetic energy of the ball when it reaches the ground = ?

Solution :-

First convert the unit of the mass of ball from gram to kilogram :

• → Mass of ball = 200 g
• → Mass of ball = (200) ÷ (1000)
• → Mass of ball = 2 ÷ 10
• → Mass of ball = 0.2 kg

Hence,the mass of the ball is 0.2 kg.

According to the Question Now :-

Finding potential energy at higest point :

• → Potential energy = mass × Acceleration due to gravity × Height
• → Potential energy = 0.2 × 9.8 × 5
• → Potential energy = 1.96 × 5
• → Potential energy = 9.8 J

Hence,the potential energy of ball is 9.8 J.

• As we know that, the Potential energy at the highest point is equal to the Kinetic energy at the lowest point.

• Hence,from the above statement we can conclude that the Kinetic energy of the ball when it reaches the ground will be 9.8 J.

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