At the Earth\'s surface, a projectile of mass 20 kgs is launched straight up at

Question

At the Earth's surface, a projectile of mass 20 kgs is launched straight up at a speed 0 kkm/s (Ignore Air resistance for this problems) To what height will it rise (using a constant value of g = 9.8 m/s2) To what height will it rise using the correct formula for potential energy as a function of distance. (Use the rigorous law of energy conservation to do this part.) Now, suppose that as it reaches its highest points, (at which it will be at rest special rocket engine to give it a new initial velocity at the new height.

Explanation / Answer

Mass of the projectile = m = 20 Kg Speed of the projectile = V = 9 Km /s = 9000 m/s Mass of the earth = ME= 6 x1024kg Radious of the earth = RE = 6.4 x106   m Gravitatinal constant = G = 6.67 x10-11 Nm2/Kg2 Using these datea we can find out , acceleration due to gravity : g = GME/RE2                                                  = (6.67 x10-11)(6 x1024)/( 6.4 x106   )2                                                  = 9.77 m/s2 (a) If we using the g value as 9.8m/s2         Height reached by the projectile : H = U2 /2g                                                                   = (9000)2/ 2 (9.8)                                                                    = 4132.65 Km Part B was not clear , If you send it clear we will helpto u Part B was not clear , If you send it clear we will helpto u

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