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The equation of projectile is y=√3x-gx×x×1÷2 horizontal range is

  1. The equation of projectile is y=√3x-gx×x×1÷2 horizontal range is
  2.  

Grade:11

2 Answers

Arun
25750 Points
5 years ago
Dear Devansh
 
Please write properly as it is notu derstandable what you have written.You can also attach an image.
 
Regards
Arun (askIITians forum expert)
Khimraj
3007 Points
5 years ago
the equation of projectile is y = √3x - gx²/2
now,
we know:-
 the equation of projectile is : y = xTanФ - gx²/ 2u²Cos²Ф
and the given equation is : y = √3x - gx²/2
on comparing both the equations ,we get

TanФ = √3
so, Ф = 60°
and 2u²Cos²Ф = 2
so u²Cos²Ф = 1
u²cos²60° = 1
u²*(1/2)² = 1
u² = 2²
so, u = 2
Hence, 2 is the initial velocity

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