Question

please answer this question​

Answer 1

Answer:

• Velocity (v) of Projection is √(g/2a).

Given:

1. Equation of Trajectory:- y = ax².

Explanation:

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We Know From Equation of Trajectory of Projectile.

$\large\bigstar \: {\boxed{\tt y = x(Tan \theta ) + \Bigg(\dfrac{g}{2u^2 Cos^2 \theta}\Bigg)x^2}}$

$\bold{Here}\begin{cases}\text{x Denotes Horizontal Distance} \\ \text{y Denotes Vertical Distance} \\ \theta \text{ Denotes Angle of projection}\end{cases}$

$\large{\boxed{\tt y = x(Tan \theta ) + \Bigg(\dfrac{g}{2u^2 Cos^2 \theta}\Bigg)x^2}}$

Comparing with the given Equation.

$\large{\tt \leadsto y = x(Tan \theta ) + \Bigg(\dfrac{g}{2u^2 Cos^2 \theta}\Bigg)x^2}$

$\large{\tt \leadsto y = ax^2}$

We get,

$\large\star \: {\tt (Tan \theta) = 0 \: -----(1)}$

$\large \star \: {\tt \dfrac{g}{2u^2 Cos^2 \theta} = a \: ----(2)}$

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Taking Equation (2).

$\large \star \:{\underline{\underline{\tt \dfrac{g}{2u^2 Cos^2 \theta} = a }}\: \tt ----(2)}$

$\large{\tt \leadsto \dfrac{g}{2u^2 Cos^2 \theta} = a}$

• We know, Cos θ = 1/Sec θ

$\large{\tt \leadsto \dfrac{g \times Sec^2 \theta}{2u^2} = a}$

• From Trigonometric identity Sec² θ = 1 + Tan² θ.

$\large{\tt \leadsto \dfrac{g \times (1 + Tan^2 \theta)}{2u^2} = a}$

Substituting Tan θ value From Equation (1).

$\large{\tt \leadsto \dfrac{g \times (1 + 0)}{2u^2} = a}$

$\large{\tt \leadsto \dfrac{g \times 1}{2u^2} = a}$

$\large{\tt \leadsto \dfrac{g}{2u^2} = a}$

$\large{\tt \leadsto g = a \times 2u^2}$

$\large{\tt \leadsto 2u^2 = \dfrac{g}{a}}$

$\large{\tt \leadsto u^2 = \dfrac{g}{2a}}$

$\large\leadsto{\underline{\boxed{\red{\tt u = \sqrt{\dfrac{g}{2a}}}}}}$

∴ Velocity (v) of Projection is √(g/2a) [Option- 4].

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