Step 1: List the data.
Mass $m = 100\ \text{g} = 0.1\ \text{kg}$, launch speed $u = 20\ \text{m s}^{-1}$, angle $\theta = 60^\circ$. We want how much kinetic energy is lost from launch up to the highest point. Take $g = 10\ \text{m s}^{-2}$.
Step 2: Kinetic energy at launch.
\[ KE_i = \tfrac{1}{2}mu^2 = \tfrac{1}{2}(0.1)(20)^2 = \tfrac{1}{2}(0.1)(400) = 20\ \text{J} \]
Step 3: What happens at the top.
At the highest point the vertical part of the velocity becomes zero, but the horizontal part stays the same the whole flight. So only the horizontal speed survives at the top. \[ v_x = u\cos 60^\circ = 20 \times 0.5 = 10\ \text{m s}^{-1} \]
Step 4: Kinetic energy at the top.
\[ KE_{top} = \tfrac{1}{2}m v_x^2 = \tfrac{1}{2}(0.1)(10)^2 = \tfrac{1}{2}(0.1)(100) = 5\ \text{J} \]
Step 5: The decrease.
\[ \Delta KE = KE_i - KE_{top} = 20 - 5 = 15\ \text{J} \]
Step 6: Conclusion.
The kinetic energy falls by 15 J during the upward journey. \[ \boxed{15\ \text{J}} \]