Civil Engineering - Applied Mechanics - Discussion

2. 

If α and u are the angle of projection and initial velocity of a projectile respectively, the horizontal range of the projectile, is

[A].
[B].
[C].
[D].

Answer: Option A

Explanation:

No answer description available for this question.

Rakesh said: (Mar 15, 2014)  
R = horizontal range = u^2 sin 2#/g.

Kaushik said: (Jul 5, 2016)  
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.

Parv said: (Jul 17, 2016)  
All options are wrong.

It will be:
= U^2 SIN2 * ANGLE/G.

Mandar said: (Nov 16, 2016)  
Agree @Parv.

Chhaya said: (Apr 4, 2017)  
I too Agree @Parv.

Anshu Singh said: (Jun 30, 2017)  
No, Given option is correct.

It should be v^2sin2θ/g.

Kishan Kumar said: (Oct 1, 2018)  
Horizontal range of the projectile, is u2sin (a)/g.
Similarly,
Toataltime of flight 2usin(a)/g.

Ajay said: (May 1, 2019)  
U^2sin2α ÷ 2g this is the correct answer.

Aswathy said: (Dec 17, 2019)  
Horizontal range= (u^2sin(2$))/g
Maximum ht=(u^2sin^2($))/2g
Time to reach maximum ht=(usin($))/g
Time of flight=(2usin($))/g

Shubham Siddharth Nayak said: (Nov 21, 2020)  
Maximum horizontal range occurs at 45 degrees.

= u sqr/g.

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