Posted by Anonymous on Thursday, March 13, 2008 at 3:28pm.
An object is projected at an angle of 53.1degrees above the horizontal at a speed of 50m/s. Sometime later it enters a narrow tube positioned at an angle of 45degrees to the vertical. Determine:
a) The initial horizontal and vertical components of velocity.
Vperp = V1Sin(theta)= 50sin53.1 = 40m/s
Vpar = V1Cos(theta)= 50cos53.1 = 30m/s
b) The horizontal and vertical velocity components as the object enters the tube.
c) The position of the mouth of the tube relative to the point from which the object was thrown.

Physics  drwls, Thursday, March 13, 2008 at 5:19pm
Your part (a) answer is correct. Part b) needs to specify where the tube is placed. It does not necessarily have to be placed where the trajectory angle is 45 degrees, but that may be what they expect you to assume. The trajectory angle is 45 degrees when Vperp = Vpar. That would mean Vperp = 30 m/s, since Vpar is fixed at that value.
c) This would involve determining the height H at which Vperp = 30 m/s.
Conservation of energy can be applied to the perpendicular portion of kinetic energy plus gravitational potential energy, so
(1/2)MVperp,0^2 = (1/2)M*(30m/s)^2 + MgH
(1/2)*40^2 = (1/2)(30^2) + gH
H = (1/2)(1600  900)/g = 35.7 m above the launch plane.
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