Question

a particle is projected with velocity u at an angle Q with the horizontal. the time after which the velocity becomes perpendicular to its initial velocity is

Answer 1

Dear Student
$$\begin{gathered} Initialvelocityofparticle=\overrightarrow{u}=u\cos \theta \hat{i}+u\sin \theta \hat{j} \\ Velocityofparticleaftertimetwhenitbecomesperpendiculartoinitialvelocity \\ \overrightarrow{v}=ucos\theta \hat{i}+\left(u\sin \theta -gt\right)\hat{j} \\ So,asbothvelocityareperpendiculartheirdotproductwillbezero. \\ \overrightarrow{u}.\overrightarrow{v}=0=\left(ucos\theta \hat{i}+usin\theta \hat{j}\right).\left(ucos\theta \hat{i}+\left(usin\theta -gt\right)\hat{j}\right) \\ {u}^2{\cos }^2\theta +u^2{\sin }^2\theta -(u\sin \theta )gt=0 \\ {u}^2(co{s}^2\theta +{\sin }^2\theta )-(usin\theta )gt=0 \\ {u}^2-(usin\theta )gt=0 \\ t=\frac{u}{g\sin \theta } \end{gathered}$$
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