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)
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)
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)
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)
V.Ram said:
6 years ago
Answer B or C? Tell me clearly.
(1)
Vim said:
5 years ago
Agree @Bazid shah.
(1)
Barqat sheikh said:
1 decade ago
Time to reach highest point = u sin30°/g.
= 9.8*0.5/9.8.
= 0.5 sec.
= 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.
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.
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