If a man weighs 50 kg and jumps to a height of 1 m what is his potential energy

The answer is as follows in the photo

The object's initial distance above the groundthe object's initial velocity

Ek₀ and ep₀ are initial kinetic and potential energies respectively.ek and ep are final kinetic and potential energies respectively.energy in the system is conserved.ek₀ + ep₀ = ek + epassume student is one some ground ground, so ep₀ = 0at the apex of student's swing, ek = 0ek₀ + 0 = 0+ epek₀ = epmv₀²/2 = mghmass cancels outv₀²/2 = ghh = v₀²/(2g)h = (3.5 m/s)²/(2 * 9.81 m/s²)h = 0.62 mhe will swing 0.62 m high (relative to where he was running before)

Let  be the velocity of the plane.let  be the velocity of the wind.let  be the resultant velocity.add the velocities to find the resultant velocity. in the attached picture, i have done the tail-to-tip method: draw one of the vectors, then add the end of the other vector to the tip of the first drawn vector.see diagram.the magnitude of  can be found via the pythagorean theorem as this is a right angle triangle.to find  , use inverse tangent: since  is in the direction "east of north", a quick convert to "north of east " reveals that 90 - 7.06 = 82.94 = 83.so our resultant is 529 m/s [83° n of e]the answer is a.