A particle is projected with a velocity u, so that its range on ahoriz

1.	A particle is projected with a velocity u, so that its range on ahorizontal plane is twice the greatest height attained. If g isacceleration due to gravity, then its range is:4g25uar2(b)(c)(d)(a) 4u25g5g5g
Answer» Given that R = 2H H = (H/2) TanθTanθ = 2 If Tanθ = 2 thenSinθ = 2 / √5Cosθ = 1 / √5 R = 2u² sinθ cosθ / g= (2u² × 2 / 5) / g= 4u² / (5 g) Range of projectile is 4u² / (5g) 0ⁿ000000ⁿπ£703.61188118xy₹% how sin is equal to √2/5 and cos is √1/5 well that is a separate mathematics question. where trignomwtric formulas are used since sinx= perpendicular/hypotenuse = 2/5 by applying pythagoras you can find that for a traingle, using pythagorean theorem you can find that, for hypotenuse =5 and perpendicular = 2 base = 1 hence cosx = base/hypotenuse=> 1/5 how you find that hypotenuse is 5

A particle is projected with a velocity u, so that its range on ahorizontal plane is twice the greatest height attained. If g isacceleration due to gravity, then its range is:4g25uar2(b)(c)(d)(a) 4u25g5g5g

Answer»

Given that R = 2H

H = (H/2) TanθTanθ = 2

If Tanθ = 2 thenSinθ = 2 / √5Cosθ = 1 / √5

R = 2u² sinθ cosθ / g= (2u² × 2 / 5) / g= 4u² / (5 g)

Range of projectile is 4u² / (5g)

0ⁿ000000ⁿπ£703.61188118xy₹%

how sin is equal to √2/5 and cos is √1/5

well that is a separate mathematics question.

where trignomwtric formulas are used

since sinx= perpendicular/hypotenuse = 2/5

by applying pythagoras you can find that

for a traingle, using pythagorean theorem you can find that, for hypotenuse =5 and perpendicular = 2

base = 1

hence cosx = base/hypotenuse=> 1/5

how you find that hypotenuse is 5

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