Discussion :: Applied Mechanics - Section 2 (Q.No.2)
If α and u are the angle of projection and initial velocity of a projectile respectively, the horizontal range of the projectile, is
Answer: Option A
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)|
|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.
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$))/g
Time to reach maximum ht=(usin($))/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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