A particle is projected with an angle of proj | vedclass

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Question

(A) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Let $k = \frac{g}{2u^2 \cos^2 	h

Vedclass · Accepted Answer

$	an ^{-1}\left[\frac{P^2+PQ+Q^2}{PQ}ight]$

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(A) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Let $k = \frac{g}{2u^2 \cos^2 	heta}$. Then the equation becomes $y = x 	an 	heta - kx^2$.Since the trajectory passes through $(P, Q)$ and $(Q, P)$,we have:$Q = P 	an 	heta - kP^2$  --- $(1)$$P = Q 	an 	heta - kQ^2$  --- $(2)$Subtracting $(2)$ from $(1)$:$Q - P = (P - Q) 	an 	heta - k(P^2 - Q^2)$$-(P - Q) = (P - Q) 	an 	heta - k(P - Q)(P + Q)$Dividing by $(P - Q)$ (assuming $P 
eq Q$):$-1 = 	an 	heta - k(P + Q) \implies k = \frac{	an 	heta + 1}{P + Q}$.Substitute $k$ into $(1)$:$Q = P 	an 	heta - \left( \frac{	an 	heta + 1}{P + Q} ight) P^2$$Q(P + Q) = P(P + Q) 	an 	heta - P^2 	an 	heta - P^2$$PQ + Q^2 + P^2 = (P^2 + PQ - P^2) 	an 	heta = PQ 	an 	heta$$	an 	heta = \frac{P^2 + Q^2 + PQ}{PQ}$.

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

$A$ particle is projected with an angle of projection $\theta$ to the horizontal. The trajectory passes through the points $(P, Q)$ and $(Q, P)$ referred to horizontal and vertical axes ($x$-axis and $y$-axis respectively). The angle of projection $\theta$ is given by:

Similar Questions

Match Column-$I$ with Column-$II$.

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

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