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Grade: 11
        
  1. The equation of projectile is y=√3x-gx×x×1÷2 horizontal range is
  2.  
11 months ago

Answers : (2)

Arun
22578 Points
							
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)
11 months ago
Khimraj
3008 Points
							
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
11 months ago
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