Question

A5kg ball is dropped from a height of 10m.

(a) Find Initial potential energy of the ball.

(b) Calculate the velocity before it reaches the ground.
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Answer 1

given,

Mass of ball(m) = 5kg

height (h) = 10 m

acceleration due to gravity(g) = 9.8 m/s^2

To find: a) Find Initial potential energy of the ball.

(b) Calculate the velocity before it reaches the ground.

Solution:

a) Potential energy is defined as the energy possessed by body by virtue of its position and configuration.

or, P.E = mgh = 5 * 9.8 * 10

or, P.E = 5 * 98 = 490 J

so, initial potential energy = 490 J

b) now, the initial potential energy at height 10 m will change into kinetic energy while reaching the ground, obeying law of conservation of energy.

or, K.E = P.E(at 10 m) = 490

and, Kinetic energy is given as:

K.E  = 1/2 mv^2

or, 490 = (1/2) * 5 * v^2

or, v² = 490*2/5

or, v² = 196

or, v = √196

or, v = 14 m/s

Therefore,the velocity before it reaches the ground is 14 m/s

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Answer 2

Given

• Mass (m) = 5 Kg
• Height (h) = 10 m
• Acceleration due to gravity (g) = 9.8 m/s²

To find

• Initial potential energy (P.E)
• Velocity before it reaches ground (K.E)

Solution

• P.E = mgh
• P.E = 5 × 9.8 × 10
• P.E = 50 × 9.8
• P.E = 490 J

Hence, the potential energy (P.E) of the ball is 490 J

Now, when the body before reaching ground, kinetic energy is possessed by the body, that is at 10 m height, and we are asked to find the velocity of the ball before it reaches the ground.

• K.E = 1/2 mv²

Since, the potential energy is transferred to kinetic energy obeying the law of conservation of energy that energy can neither be created nor destroyed, but only transferred from one form to another.

Hence, P.E = K.E :-

• 490 = 1/2 mv²
• 490 = 1/2 5v²
• v² = 490 × 2/5
• v² = 196
• v = √196
• v = 14

Hence, the velocity before the ball reaches the ground is 14 m/s