The trajectory of a projectile in a vertical | vedclass

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Question

(A) The given equation of the trajectory is $y = \alpha x - \beta x^2$. Comparing this with the standard equation of a projectile trajectory: $y = x \

Vedclass · Accepted Answer

$	an^{-1} \alpha, \frac{\alpha^2}{4\beta}$

Vedclass · Answer

(A) The given equation of the trajectory is $y = \alpha x - \beta x^2$. Comparing this with the standard equation of a projectile trajectory: $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$. By comparing the coefficients of $x$,we get $	an 	heta = \alpha$,which implies $	heta = 	an^{-1} \alpha$. By comparing the coefficients of $x^2$,we get $\beta = \frac{g}{2u^2 \cos^2 	heta}$. Rearranging for $u^2$,we find $u^2 = \frac{g}{2\beta \cos^2 	heta}$. The formula for maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$. Substituting $u^2$ into the formula: $H = \frac{g}{2\beta \cos^2 	heta} \cdot \frac{\sin^2 	heta}{2g} = \frac{	an^2 	heta}{4\beta}$. Since $	an 	heta = \alpha$,we have $H = \frac{\alpha^2}{4\beta}$.

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