A projectile is launched from the origin in t | vedclass

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

Question

(B) The distance $r$ from the origin is given by $r^2 = x^2 + y^2$. For $r$ to increase monotonically with $x$,we require $\frac{dr}{dx} > 0$,which is

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{1}{3}\right)$

Vedclass · Answer

(B) The distance $r$ from the origin is given by $r^2 = x^2 + y^2$. For $r$ to increase monotonically with $x$,we require $\frac{dr}{dx} > 0$,which is equivalent to $\frac{d(r^2)}{dt} > 0$ since $x$ increases with time $t$.Given $x = v_0 \cos \alpha \cdot t$ and $y = v_0 \sin \alpha \cdot t - \frac{1}{2}gt^2$,we have:$r^2 = (v_0 \cos \alpha \cdot t)^2 + (v_0 \sin \alpha \cdot t - \frac{1}{2}gt^2)^2$$r^2 = v_0^2 t^2 \cos^2 \alpha + v_0^2 t^2 \sin^2 \alpha - v_0 \sin \alpha \cdot g t^3 + \frac{1}{4}g^2 t^4$$r^2 = v_0^2 t^2 - v_0 \sin \alpha \cdot g t^3 + \frac{1}{4}g^2 t^4$Differentiating with respect to $t$:$\frac{d(r^2)}{dt} = 2 v_0^2 t - 3 v_0 \sin \alpha \cdot g t^2 + g^2 t^3$For $r$ to increase,$\frac{d(r^2)}{dt} > 0$ for all $t > 0$. Dividing by $t$:$g^2 t^2 - 3 v_0 \sin \alpha \cdot g t + 2 v_0^2 > 0$This quadratic in $t$ is always positive if its discriminant $D $D = (-3 v_0 \sin \alpha \cdot g)^2 - 4(g^2)(2 v_0^2) $9 v_0^2 g^2 \sin^2 \alpha - 8 v_0^2 g^2 $\sin^2 \alpha Since $\cos^2 \alpha = 1 - \sin^2 \alpha$,we have $\cos^2 \alpha > 1 - \frac{8}{9} = \frac{1}{9}$.Thus,$\cos \alpha > \frac{1}{3}$. The critical angle is $\alpha_c = \cos^{-1}\left(\frac{1}{3}\right)$.

Similar Questions

$A$ body of mass $1 \, kg$ starts moving from rest at $t = 0$ in a circular path of radius $8 \, m$. Its kinetic energy varies as a function of time as: $K.E. = 2t^2 \, J$,where $t$ is in seconds. Then:

$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then the product of the two times of flight is proportional to

Four bodies $P, Q, R$ and $S$ are projected with equal velocities having angles of projection $15^{\circ}, 30^{\circ}, 45^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The body having the shortest range is

$A$ ball of mass $100 \ g$ is projected with velocity $20 \ m/s$ at $60^{\circ}$ with the horizontal. The decrease in kinetic energy of the ball during the motion from the point of projection to the highest point is:

$A$ body moves along a circular path of radius $100 \,m$ and completes one revolution in $40 \,sec$. What is the total distance covered at the end of $2 \,min \,20 \,sec$ (in $\pi$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module