Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 3 (Q.No. 9)
9.
A jet projected at an angle of 45θ, 40 m from the foot of a vertical column, just reaches the top of the column. The height of the column is
Discussion:
14 comments Page 1 of 2.
Rakesh roushan said:
3 years ago
Agree with you, thanks @Pkota.
(1)
Pkota said:
4 years ago
Projectile motion.
Range R=u^2sin2θ/g.
Height H=u^2(sinθ)^2/2g,
Here 40m is R/2 so R=80,
i.e.R=u^2sin2(45°)/g ; 80=u^2/g.
H = u^2 (sin45°)^2/2g =40* (1/√2)^2=20m.
Range R=u^2sin2θ/g.
Height H=u^2(sinθ)^2/2g,
Here 40m is R/2 so R=80,
i.e.R=u^2sin2(45°)/g ; 80=u^2/g.
H = u^2 (sin45°)^2/2g =40* (1/√2)^2=20m.
(2)
Anjaligangwar said:
5 years ago
tan45° = height of the column/distance between jet and column (By Pythagoras theorem).
Gege said:
7 years ago
From any point jet has departed 45° and reaches the height of column (H).
The distance between from any point to the tip of the column is 40m.
This can be calculated using " Angle of Elevation"
Tan(45°) = H/40;
H = 40Tan(45°).
H = 40m.
The distance between from any point to the tip of the column is 40m.
This can be calculated using " Angle of Elevation"
Tan(45°) = H/40;
H = 40Tan(45°).
H = 40m.
(4)
Viplav said:
7 years ago
40=v^2sin(90)/g.
V=20.
H=20^2 sin(45)^2/2g=15.
Answer is 15.
V=20.
H=20^2 sin(45)^2/2g=15.
Answer is 15.
Nikhil said:
7 years ago
40tan45°= 40m.
Priya said:
7 years ago
Horizontal range 40*2=u^2sin2θ/g. We get u=28.
Hmax=u^2sin(θ)^2/2g Hmax=20m.
Hmax=u^2sin(θ)^2/2g Hmax=20m.
Jony said:
8 years ago
Please derive the expression @Garry.
Garry said:
8 years ago
The path of the particle will be a parabolic so tan θ will not be applied the correct answer from the projectile motion will come to be 20 m.
And I also agree @Mahato.
And I also agree @Mahato.
Krishan Suthar said:
9 years ago
Yes, Correct @Aditya.
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