To apply the law of conservation of energy to an object launched upward in Earth

Question

To apply the law of conservation of energy to an object launched upward in Earth's gravitational field. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K=(1/2)mv^2 and its gravitational potential energy U=mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation K_{ m i}+U_{ m i}=K_{ m f}+U_{ m f};;;;, where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. As the projectile goes upward, what energy changes take place?

Explanation / Answer

1) v = initial velocity, V = velocity at any instance V = v - gt 1/2 v = v - gt gt = v -1/2v t = v/2g h = vt - 1/2 gt^2 h = v(v/2g) - 1/2g(v/2g)^2 h = (v^2) / 2g - (v^2) /8g h = (3/8) (v^2)/g This is the height at 0.5v 2) There should be two answers, one going up and the other coming down. Same value, just different sign or directions. h = vt - 1/2 gt^2 if h = 1/2 h_max or h = 1/2 ((v^2)/2g) = (v^2)/4g (v^2)/4g = vt - 1/2 gt^2 solve for t (v^2)/4g - vt + 1/2 gt^2 = 0 (v^2)/2g - 2vt + gt^2 = 0 (multiply by 2) gt^2 - 2vt + (v^2)/2g = 0 (rearrange) Use quadratic formula: t = { --2v +/- SQRT[ (2v)^2 - 4 g (v^2)/2g] / 2g} do some algebra to reduce to t = { 2v +/- SQRT[2] v} /2g = (1 +/- 1.414/2) (v/g) u = v - gt (Original velocity equation) u = v - g (1 +/- 1.414/2) (v/g) u = v ( +/- 1.414/2) u = +/- 0.707 v

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