Can someone solve for the KE, PE, AND TE for both trails of each point, please and thankyou. I don't understand how to calculate.

Data Table 2: Trajectory Points for Trial 5 Trial 5: Point on Position Velocity PE KE TE trajector (m) (m/s) () (J) () y PE= mgh KE=1/2mv2 TE= PE+ KE 0.683m 1.806 m/s 9.85 J 0.24 J 10.1 J Point 1 0.818m 0.869 m/s 11.97J 0.06 J 12.0 J Point 2 0.856m -0.092 12.34 J 0.0 J 12.3 J Point 3 m/s 0.799m -1.026 11.51 J 0.08 J 11.6 J Point 4 m/s 0.651m -1.895 9.38 J 0.27 J 9.7 J Point 5 m/sData Table 3: Trajectory Points for Trial 7 Trial 7: Point on Position Velocity PE KE TE trajector (m) (m/s) (J) () () y PE=mgh KE=1/2mv TE= PE+ KE 0.454 m 1.733 m/s 6.54 J 0.23 J 6.8J Point 1 0.589 m 0.869 m/s 8.49 J 0.06 J 8.6 Point 2 0.628 m -0.063 9.05 J 0.0 J 9.05J Point 3 m/s 0.575 m -1.002 8.29 J 0.08 J 8.37J Point 4 m/s 0.428 m -1.709 6.17 J 0.22 J 6.39] Point 5 m/s\f1. Potential Energy (PE)= mgh. where , m = mass in kg g= acceleration due to gravity. h= height in. meter 2. Kinetic energy (KE)= 1/2 mv^2 where, m= mass in kilograms v= velocity 3. Total Energy (TE) = Potential energy (PE) + kinetic energy (KE) TE= mgh+1/2 mv^2.Analysis: 1. Calculate PE, KE, and TE for all points in both trials