A ball of mass 100 g is projected with velocity 20 m/s at$60°$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is :

Step 1: Initial Kinetic Energy (K.E.)

The total kinetic energy at the point of projection is given by:

$$\text{K.E.}_\text{initial} = \frac{1}{2} m u^2$$

Substitute the given values:

$$\text{K.E.}_\text{initial} = \frac{1}{2} (0.1) (20)^2 = 20 \, \text{J}.$$

Step 2: Velocity at the Highest Point

At the highest point, the vertical component of the velocity becomes zero ($v_y = 0$), and only the horizontal component of the velocity ($v_x$) remains. The horizontal velocity is:

$$v_x = u \cos\theta$$

Substitute the values:

$$v_x = 20 \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s}.$$

Step 3: Kinetic Energy at the Highest Point

The kinetic energy at the highest point is due to the horizontal velocity only:

$$\text{K.E.}_\text{highest} = \frac{1}{2} m v_x^2$$

Substitute the values:

$$\text{K.E.}_\text{highest} = \frac{1}{2} (0.1) (10)^2 = 5 \, \text{J}.$$

Step 4: Decrease in Kinetic Energy

The decrease in kinetic energy during the motion from the point of projection to the highest point is:

$$\Delta \text{K.E.} = \text{K.E.}_\text{initial} - \text{K.E.}_\text{highest}$$

Substitute the values:

$$\Delta \text{K.E.} = 20 - 5 = 15 \, \text{J}.$$