Civil Engineering - Hydraulics - Discussion

Discussion Forum : Hydraulics - Section 2 (Q.No. 11)
11.
If jet of water coming out from a nozzle with a velocity 9.81 m/s, the angle of elevation being 30°, the time to reach the highest point is
0.25 s
0.50 s
1.0 s
1.5 s.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Barqat sheikh said:   1 decade ago
Time to reach highest point = u sin30°/g.

= 9.8*0.5/9.8.

= 0.5 sec.

Srinu said:   9 years ago
Thanks @Barqat Sheikh.

Dhvajvahak said:   9 years ago
Correct ans is option C, 1 sec T = 2 V sin30/g.

Huud said:   8 years ago
s = v.tsin30,
s = 1/2 * gt2.

Then vtsin30 = 1/2gt2.
9.81m/s=gt since your cancel sin30 = 1/2.
t = 1sec so correct answer is c.
(1)

Mhathung kikon said:   8 years ago
TIME TO FLIGHT.

This is the time to complete the projectile path which is 2 x (time to reach the maximum height).
T = 2 x USinθ/g.
so the correct answer is C.

Kishor said:   8 years ago
t=2 u sinθ/g is total time duration but we need highest point reach time then t=u sinθ/g.

Anoop Boora said:   8 years ago
Time to reach highest point = u sinθ/g.
Time of flight = 2 x using /g.
Horizontal range of jet = u^2 sin2θ/g.
(1)

Nilraj said:   7 years ago
Thanks @Anoop.

V.Ram said:   6 years ago
Answer B or C? Tell me clearly.
(1)

Bazid shah said:   6 years ago
Time of flight = 2usinθ/g.

As we are calculating the time taken to reach the highest point so we will divide it by 2 because in the first half time the jet will go upwards then it will drop. So the answer will be 0.5.
(2)


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