If at any point on the path of a projectile i | vedclass

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

Question

(B) Let the initial velocity be $u$ at an angle $\alpha$ with the horizontal. The horizontal component is $u_x = u \cos \alpha$ and the vertical compo

Vedclass · Accepted Answer

$\frac{u}{g \sin \alpha}$

Vedclass · Answer

(B) Let the initial velocity be $u$ at an angle $\alpha$ with the horizontal. The horizontal component is $u_x = u \cos \alpha$ and the vertical component is $u_y = u \sin \alpha$.After time $t$,let the velocity be $v$. The horizontal component remains constant: $v_x = u \cos \alpha$.The vertical component becomes $v_y = u \sin \alpha - gt$.For the velocity to be at right angles to the initial direction,the dot product of the initial velocity vector $\vec{u}$ and the final velocity vector $\vec{v}$ must be zero,or more simply,the slope of the new velocity must be the negative reciprocal of the initial slope.Alternatively,using the condition that the velocity vector $\vec{v}$ is perpendicular to $\vec{u}$,we have $\vec{u} \cdot \vec{v} = 0$.$(u \cos \alpha \hat{i} + u \sin \alpha \hat{j}) \cdot (u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j}) = 0$$u^2 \cos^2 \alpha + u \sin \alpha (u \sin \alpha - gt) = 0$$u^2 \cos^2 \alpha + u^2 \sin^2 \alpha - ugt \sin \alpha = 0$$u^2 (\cos^2 \alpha + \sin^2 \alpha) = ugt \sin \alpha$$u^2 = ugt \sin \alpha$$t = \frac{u}{g \sin \alpha}$

If at any point on the path of a projectile its velocity is $u$ at an inclination $\alpha$ to the horizontal,then after what time will it move at right angles to its former direction?

Similar Questions

$A$ particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t)=A[\hat{i} \cos (kt)-\hat{j} \sin (kt)]$,where $A$ and $k$ are constants. The angle between the force and the momentum is (in $^{\circ}$)

$A$ body is projected at $t=0$ with a velocity of $10 \ m/s$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t=1 \ s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g=10 \ m/s^2$, the value of $R$ is: (in $m$)

$A$ car is moving with a velocity of $4 \,m \,s^{-1}$ towards east. After a time of $4 \,s$, if it is heading north-east with a velocity of $4 \sqrt{2} \,m \,s^{-1}$, then the average velocity of the car is

Find the speed of the intersection point $O$ of the two wires if the wires start moving perpendicular to themselves with speed $v$ as shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module