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.
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.
To determine the horizontal and vertical velocity components as the object enters the tube, we need to consider the angle of the tube relative to the vertical.
Let's assume the horizontal velocity component remains the same because there is no force acting on it. So, the horizontal velocity component as the object enters the tube would still be 30 m/s.
Now, for the vertical velocity component, we need to take into account the angle of the tube. The tube is positioned at an angle of 45 degrees to the vertical.
To find the vertical velocity component as the object enters the tube, we can use the equation: Vtube = Vperp*cos(theta).
Here, Vperp is the initial perpendicular velocity component, which we found to be 40 m/s, and theta is the angle of the tube, which is 45 degrees.
Vtube = 40 * cos(45)
Vtube = 40 * 0.7071
Vtube ≈ 28.284 m/s
So, the vertical velocity component as the object enters the tube is approximately 28.284 m/s.
To determine the position of the mouth of the tube relative to the point from which the object was thrown, we need to know the time it takes for the object to enter the tube.
Unfortunately, the information provided in the question does not include the time it takes for the object to reach the tube. Without this information, we cannot determine the precise position of the mouth of the tube relative to the point of projection.