Path of projectile is given by the equation Y | vedclass

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

Question

(A) The equation of the trajectory is $Y = P x - Q x^2$.Comparing this with the standard equation $Y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	het

Vedclass · Accepted Answer

$(A)-(i)$,$(B)-(iii)$,$(C)-(iv)$,$(D)-(ii)$

Vedclass · Answer

(A) The equation of the trajectory is $Y = P x - Q x^2$.Comparing this with the standard equation $Y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$, we get $	an 	heta = P$ and $Q = \frac{g}{2 u^2 \cos^2 	heta}$.$(A)$ Range $(R)$: At $Y = 0$, $x = R$. So, $0 = P R - Q R^2 \Rightarrow R = \frac{P}{Q}$. Thus, $(A)-(i)$.$(B)$ Maximum height $(H)$: $H = \frac{R 	an 	heta}{4} = \frac{(P/Q) \cdot P}{4} = \frac{P^2}{4 Q}$. Thus, $(B)-(iii)$.$(C)$ Time of flight $(T)$: $T = \frac{2 u \sin 	heta}{g}$. Since $	an 	heta = P$ and $Q = \frac{g}{2 u^2 \cos^2 	heta}$, we have $u \cos 	heta = \sqrt{\frac{g}{2 Q}}$. Then $T = \frac{2 u \sin 	heta}{g} = \frac{2 (u \cos 	heta) 	an 	heta}{g} = \frac{2 \sqrt{\frac{g}{2 Q}} \cdot P}{g} = \left(\sqrt{\frac{2}{g Q}}ight) P$. Thus, $(C)-(iv)$.$(D)$ Tangent of projection: $	an 	heta = P$. Thus, $(D)-(ii)$.

Similar Questions

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $150\, ms^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is

$A$ projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40\,m/s$. The velocity of the projectile at $t = 2\,s$ from the start will be $........$ (Given $g = 10\,m/s^2$)

$A$ projectile is given an initial velocity of $(\hat i + 2\hat j) \ ms^{-1}$,where $\hat i$ is along the ground and $\hat j$ is along the vertical. If $g = 10 \ m/s^2$,the equation of its trajectory is

$A$ body is thrown at an angle of $30^{\circ}$ to the horizontal with a velocity of $30\; m/s$. After $1\; s$,its velocity will be (in $m/s$) $\left(g=10\; m/s^{2}\right)$

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal from a distance $d_1$ from the foot of a pole. It just touches the top of the pole and then falls on the ground at a distance $d_2$ from the pole on the other side. The height of the pole is ...........

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$(A)$ Range	$(i)$ $\frac{P}{Q}$
$(B)$ Maximum height	$(ii)$ $P$
$(C)$ Time of flight	$(iii)$ $\frac{P^2}{4 Q}$
$(D)$ Tangent of projection	$(iv)$ $\left(\sqrt{\frac{2}{g Q}}\right) P$