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UPSEE 2019 Solved Paper

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Question : 3 of 150

Marks: +1, -0

The trajectory of a projectile in a vertical plane is

y=\alpha x-\beta x^{2}

\alpha

\beta

are constants and

x

y

are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attended by projectile is

$\frac{\alpha}{\beta}$
$\frac{\alpha^{2}}{4 \beta}$
$\frac{\beta}{\alpha}$
$\frac{\beta}{2 \alpha}$

Solution:

Here, the trajectory of a projectile motion,

y=\alpha x-\beta x^{2}

.......(i)

As, general equation of trajectory of a projectile motion,

y=\tan \theta \cdot x-\frac{g}{2(u \cos \theta)^{2}} \cdot x^{2}

........(ii)

Now, comparing Eqs. (i) and (ii), we get

\tan \theta = \alpha \Rightarrow \frac{\sin^{2} \theta}{\cos^{2} \theta} = \alpha^{2}

........(iii)

and

\beta = \frac{g}{2 u^{2} \cos^{2} \theta}

\Rightarrow 4 \beta = \frac{2 g}{u^{2} \cos^{2} \theta}

........(iv)

Maximum height of projectile,

H_{\max} = \frac{u^{2} \sin^{2} \theta}{2 g}

From Eqs. (iii) and (iv), we get

\frac{\alpha^{2}}{4 \beta} = \frac{u^{2} \cos^{2} \theta \times \sin^{2} \theta}{2 g \cos^{2} \theta} = H_{\max}

Hence, the maximum height is

\frac{\alpha^{2}}{4 \beta}

.

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