January 19, 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.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions