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
Discussion:
12 comments Page 1 of 2.
Jero M said:
2 years ago
Time to reach the highest point,
t= uSin(θ)/g = 9.81 x Sin30°/9.81 = 0.5.
t= uSin(θ)/g = 9.81 x Sin30°/9.81 = 0.5.
(5)
Vim said:
5 years ago
Agree @Bazid shah.
(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.
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)
V.Ram said:
6 years ago
Answer B or C? Tell me clearly.
(1)
Nilraj said:
7 years ago
Thanks @Anoop.
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.
Time of flight = 2 x using /g.
Horizontal range of jet = u^2 sin2θ/g.
(1)
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.
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.
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.
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.
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)
Dhvajvahak said:
9 years ago
Correct ans is option C, 1 sec T = 2 V sin30/g.
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