A particle is projected with an angle of proj | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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]$

Vedclass · Answer

(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}$.

$A$ 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

$A$ particle aimed at a target,projected with an angle $15^{\circ}$ with the horizontal,is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$,it is away from the target by $15 \ m$. Then the angle of projection to hit the target is:

$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$, the maximum height reached by it is: (in meters)

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

The equation of motion of a projectile is $y = 12x - \frac{3}{4}x^2$. The range of the projectile is $..........\,m$.

$A$ body projected from the ground reaches a point $X$ in its path after $3 \ s$ and from there it reaches the ground after a further $6 \ s$. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module