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

A ball of mass $$m = 100 \text{ g} = 0.1 \text{ kg}$$ is projected with velocity $$u = 20 \text{ m/s}$$ at $$\theta = 60°$$ with the horizontal.

Since the kinetic energy at the point of projection is $$KE_i = \frac{1}{2}mu^2 = \frac{1}{2} \times 0.1 \times 20^2 = \frac{1}{2} \times 0.1 \times 400 = 20 \text{ J}$$

At the highest point of projectile motion, the vertical component of velocity becomes zero and only the horizontal component remains. From this, the velocity at the highest point is $$v = u\cos\theta = 20 \times \cos 60° = 20 \times \frac{1}{2} = 10 \text{ m/s}$$

Next, the kinetic energy at the highest point is $$KE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.1 \times 10^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \text{ J}$$

Finally, the decrease in kinetic energy is calculated as $$\Delta KE = KE_i - KE_f = 20 - 5 = 15 \text{ J}$$

The correct answer is Option (2): $$\boxed{15 \text{ J}}$$.