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.

I answered this about two days ago. Please check previous postings. If you need additional help, show your work.

thanks! Is the answer supposed to be -30m/s for Vperp, though? Since it's going down into the tube?

Also I don't know how to find time without distance.
The only way to do it can be
(V2-V1)/a
30-40/-10
(*we are supposed to use 10 for acceleration)
= 1s
I'm not sure if that's right.

d)
Would you multiply the perpendicular component of V2 by the time found for the height?

there are of course two possible solutions. It could be at 45 degrees on the way up or on the way down.

Your comment that it goes DOWN into the tube means that indeed your vertical component is - 30 while your horizontal component remains +30
Now
v = vertical speed = Vo - g t
so
-30 = 40 - 10 t
10 t = 70
t = 7 seconds to tube entry

Now where did it go in those seven seconds
x = 30 t = 30 * 7 = 210 meters
h = height = Vo t + (1/2) a t^2
but a = -10 m/s^2
h = 40*7 -5*49
h = 280-245
h = 35 meters

hi

1tdc03sz4jr9fith
good luck

c) To determine the time taken to reach the tube, we need to find the horizontal distance traveled by the object. We can use the initial horizontal velocity and the time taken to reach the tube.

The formula to calculate the horizontal distance is: Distance = Velocity x Time.

Since the initial horizontal velocity is 30 m/s, we can rearrange the formula to solve for time:
Time = Distance / Velocity.

Given that the distance is unknown, we can determine it by calculating the vertical distance using the initial vertical velocity and the time taken to reach the tube.

The formula to calculate the vertical distance is: Distance = Velocity x Time + 0.5 x Acceleration x Time^2.

In this case, we know that the acceleration due to gravity is acting vertically downwards and its value is -9.8 m/s^2. The initial vertical velocity is 40 m/s (calculated previously). Since we are looking for the time taken to reach the tube, the vertical distance traveled should be zero (since the object will be at the same height when it enters the tube).

0 = 40 x Time + 0.5 x (-9.8) x Time^2.

This equation is a quadratic equation that we can solve to find the time(s) at which the above equation is satisfied.

d) To determine the position of the mouth of the tube relative to the point from which the object was thrown, we need to find the horizontal and vertical distances traveled by the object.

The horizontal distance is the same as the distance calculated previously using the time.

The vertical distance can be found using the equation used before: Distance = Velocity x Time + 0.5 x Acceleration x Time^2.

Since we know the time, we can substitute it into the equation to find the vertical distance.

Finally, to determine the position relative to the initial point, we can use the Pythagorean theorem: Distance^2 = (Horizontal Distance)^2 + (Vertical Distance)^2.