A shell is shot with an initial velocity of 26 m/s, at an angle of ?0 = 63? with
ID: 2146643 • Letter: A
Question
A shell is shot with an initial velocity of 26 m/s, at an angle of ?0 = 63? with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-42). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?
A shell is shot with an initial velocity of 26 m/s, at an angle of ?0 = 63? with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-42). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?Explanation / Answer
sol i perform earlier with diff values plug your values it helps initial shell at max height, v=0 v2=u2-2gs 0 = (20 sin 50)^2 -2 *9.81* s s = 11.96m (max height) v=u -gt 0 = 20 sin 50 - 9.81 *t t = 1.56 s horizontal dis at max height = 20 cos 50 * 1.56 = 20.06 m At the top of the trajectory, vy=0 , vx = 20 cos 50 conservation of momentum m 20 cos 50 = 0.5 m *0 + 0.5 m * u 20 cos 50 = 0.5 u u = 25.7 m/s (horizontally, speed of other part) vertical motion s = ut + 0.5 gt^2 11.96 = 0 +0.5 * 9.81* t^2 t = 1.56 s horizontal dis from max height = 25.7 * 1.56 = 40.1 m so total distance the other fragment lands = 20.06 + 40.1 = 60.16 m ---amswer
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