Question

A ball of mass 1 kg is thrown vertically upward with speed of 30 m/s. Find the potential energy when it is at half of the maximum height. [Take g = 10 m/s2 ]



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Answer 1

Given :

Mass of ball = 1 kg

Initial velocity = 30 m/s

Acc. due to gravity = 10 m/s²

To Find :

Potential energy of ball when it is at half of the maximum height.

Solution :

❒ For a body thrown vertically upward, g is taken negative.

First of all we need to find maximum height attained by the ball$.$

Third equation of kinematics is given by

• v² - u² = 2gH

» v denotes final velocity

» u denotes initial velocity

» g denotes acceleration

» H denotes height

At maxima height, v = 0

By substituting the given values;

➙ 0² - 30² = 2(-10)H

➙ -900 = -20H

➙ H = 900/20

➙ H = 45 m

We are asked to potential energy at half of the maximum height. i.e., at 22.5 m

➙ U = m g h

➙ U = 1 × 10 × 22.5

➙ U = 225 J