Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 2 (Q.No. 2)
2.
If α and u are the angle of projection and initial velocity of a projectile respectively, the horizontal range of the projectile, is
Discussion:
10 comments Page 1 of 1.
SHUBHAM SIDDHARTH NAYAK said:
5 years ago
Maximum horizontal range occurs at 45 degrees.
= u sqr/g.
= u sqr/g.
Aswathy said:
6 years ago
Horizontal range= (u^2sin(2$))/g
Maximum ht=(u^2sin^2($))/2g
Time to reach maximum ht=(usin($))/g
Time of flight=(2usin($))/g
Maximum ht=(u^2sin^2($))/2g
Time to reach maximum ht=(usin($))/g
Time of flight=(2usin($))/g
Ajay said:
6 years ago
U^2sin2α ÷ 2g this is the correct answer.
Kishan kumar said:
7 years ago
Horizontal range of the projectile, is u2sin (a)/g.
Similarly,
Toataltime of flight 2usin(a)/g.
Similarly,
Toataltime of flight 2usin(a)/g.
Anshu singh said:
8 years ago
No, Given option is correct.
It should be v^2sin2θ/g.
It should be v^2sin2θ/g.
Chhaya said:
8 years ago
I too Agree @Parv.
MANDAR said:
9 years ago
Agree @Parv.
PARV said:
9 years ago
All options are wrong.
It will be:
= U^2 SIN2 * ANGLE/G.
It will be:
= U^2 SIN2 * ANGLE/G.
Kaushik said:
9 years ago
Answer should be U^2sin(2a)/g
As range will be horizontal component of velocity i.e Ux multiply by t(time)____ Ux t= Usin(a)*t
now for finding t let us consider vertical motion of proectile
So, S = ut + 0.5 at^2.
0 = Uy *t - 0.5 * a * t^2 (put Uy as Usin(a)).
Solving this we get t = 2Usin(a)/g.
Hence range R = Ux * t = Ucos(a) * 2 * Usin(a)/g.
Hence R = U^2sin(2a)/g.
As range will be horizontal component of velocity i.e Ux multiply by t(time)____ Ux t= Usin(a)*t
now for finding t let us consider vertical motion of proectile
So, S = ut + 0.5 at^2.
0 = Uy *t - 0.5 * a * t^2 (put Uy as Usin(a)).
Solving this we get t = 2Usin(a)/g.
Hence range R = Ux * t = Ucos(a) * 2 * Usin(a)/g.
Hence R = U^2sin(2a)/g.
RAKESH said:
1 decade ago
R = horizontal range = u^2 sin 2#/g.
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