Question

A 20kg ball is thrown upwards with a speed of 6m/second find its P.E when it reaches the highest point also find the maximum height it reaches (g=10m/ second ^2

Answer 1

Answer:

Given:

Mass of ball = 20 kg

Initial Velocity = 6 m/s

To find:

Potential Energy of the object at the highest point.

Concept:

First we will find the height to which the object rises with the help of Equations of Kinematics.

Then we will find the potential energy.

Calculation:

v² = u² + 2ah

=> 0² = 6² + 2 × (-10) × h

=> 20h = 36

=> h = 36/20

=> h = 1.8 metres.

Potential Energy at that height :

PE = mgh

=> PE = 20 × 10 × (1.8)

=> PE = 360 Joules.

So final answer is 360 Joules

Answer 2

Explanation:-

Given that:-

• Weight (mass) of the ball = 20kg .

• Speed of ball = 6m/s.

To find:-

• Potential energy ( P.E) when the ball reaches at the highest point.

• Maximum height it reaches.

____________________________

Solution:-

$\implies$$\tt{v {}^{2} = v {}^{2} + 2ah}$

$\implies$$\tt{0 {}^{2} = 6 {}^{2} + 2 \times ( - 10) \times h}$

$\implies$ $\tt{20h = 36}$

$\implies$$\tt{h = \frac{36}{20}}$

$\implies$$\tt{h = 1.8m}$

Therefore, the maximum height = 1.8m

Now, we will find Potential energy=

$\implies$$\tt{P.E = mgh}$

$\implies$$\tt{P.E = 20 \times 10 \times (1.8)}$

$\implies$$\tt{P.E = 360 \: Joules}$

Therefore, Potential energy = 360 joules.