# Physics

posted by on .

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
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.

• Physics - ,

• Physics - ,

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?

• Physics - ,

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

• Physics - ,

hi
1tdc03sz4jr9fith
good luck