1. **Problem statement:** Calculate the kinetic energy of a dart hitting a target at a speed of 4.39 m/s, then find the height from which the dart must be dropped to reach that energy level without air resistance. 2. **Given data:** - Speed $v = 4.39$ m/s - Mass $m = 0.2$ kg - Gravitational acceleration $g = 9.81$ m/s$^2$ - Height $h$ to be found 3. **Step 1: Calculate kinetic energy (KE) when dart hits the target.** Formula for kinetic energy: $$ KE = \frac{1}{2} m v^2 $$ Substitute values: $$ KE = \frac{1}{2} \times 0.2 \times (4.39)^2 $$ Calculate: $$ KE = 0.1 \times 19.2721 = 1.92721 \text{ J} $$ 4. **Step 2: Calculate height $h$ from which dart must be dropped to have the same energy (potential energy equals kinetic energy).** Formula for potential energy: $$ PE = m g h $$ Set $PE = KE$: $$ m g h = KE $$ Solve for $h$: $$ h = \frac{KE}{m g} $$ Substitute values: $$ h = \frac{1.92721}{0.2 \times 9.81} $$ Show cancellation: $$ h = \frac{1.92721}{\cancel{0.2} \times 9.81} = \frac{1.92721}{1.962} $$ Calculate: $$ h = 0.9825 \text{ meters} $$ 5. **Final answers:** - Kinetic energy at impact: $1.92721$ J - Height to drop dart for same energy: $0.9825$ m These values can be used to determine coordinates as per the problem instructions.