Question

Q1 A ball rolls off the top of a table 1.2m high and lands on the floor at a horizontal distance of 2m. What was the velocity as it left the table?

Answer 1

Answer:

it's velocity is aprox 5m/ sec

Answer 2

Velocity of ball when it left the table is √23.52 m s⁻¹

Explanation:

Given, height of table = 1.2 m

horizontal distance = 2 m

Let the mass of ball = m

velocity when it left the table = v

acceleration due to gravity = g= 9.8

• When body is reached at some height it possess potential energy and when it moves with some velocity it possess kinetic energy

                             Potential Energy = m g h

                              Kinetic Energy = $\frac{1}{2}$ m v²

• As body is raised up to a height so it have potential energy when it start moving it have kinetic energy, when body left the table its potential energy gets converted into kinetic energy so the potential energy will be equal to kinetic energy So, Potential energy = Kinetic energy

=> m g h = $\frac{1}{2}$ m v²

=> g h = $\frac{1}{2}$ v²

=> v² =  2 g h

=> v = √2 g h

=> v = √2* 9.8 *1.2

=> v = √ 9.8 *2.4

=> v = √23.52 m s⁻¹

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