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

Arun
22578 Points

Dear Devansh

Please write properly as it is notu derstandable what you have written.You can also attach an image.

Regards
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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• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions