The equation of the trajectory of a ball proj | vedclass

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(C) The standard equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the

Vedclass · Accepted Answer

$\sqrt{2} 	ext{ m/s}$

Vedclass · Answer

(C) The standard equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing this with the given equation $y = x - \frac{gx^2}{2}$, we get $	an 	heta = 1$, which implies $	heta = 45^{\circ}$.Also, by comparing the coefficients of $x^2$, we have $\frac{g}{2u^2 \cos^2 	heta} = \frac{g}{2}$.This simplifies to $u^2 \cos^2 	heta = 1$.Substituting $	heta = 45^{\circ}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$, we get $u^2 (\frac{1}{\sqrt{2}})^2 = 1$.$u^2 (\frac{1}{2}) = 1 \implies u^2 = 2$.Therefore, the initial velocity $u = \sqrt{2} 	ext{ m/s}$.

Column-$I$	Column-$II$
$(1)$ Vertical component of velocity is zero	$(a)$ Tangent to the parabolic path
$(2)$ Linear velocity	$(b)$ Maximum height point of the projectile trajectory

The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal is given as $y = x - \frac{gx^2}{2}$. The initial velocity of the ball is $[\text{Given} : \tan 45^{\circ} = 1, \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$

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Column-$I$ Column-$II$

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