February 27, 2017

Homework Help: Physics

Posted by Anonymous on Friday, March 14, 2008 at 3:16pm.

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.
Since Vhorizontal stays the same it will be 30m/s.
Then I did
Hyp = Adjacent/Cos(theta)
and got = 42.43
Vvertical = V1Sin(theta) = 42.43Sin(45) = 30m/s

c) the time taken to reach the tube

d) the position of the mouth of the tube relative to the point from which the object was thrown.

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