A projectile is fired from the origin with an initial velocity $v_0$ at an angle $\theta$ with the horizontal. At what time will the velocity vector be perpendicular to the initial velocity vector?

Solution

1. The initial velocity is ${\vec{v}}_0=(v_0\cos \theta )\hat{i}+(v_0\sin \theta )\hat{j}$, and the velocity at time $t$ is $\vec{v}=(v_0\cos \theta )\hat{i}+(v_0\sin \theta -gt)\hat{j}$.
2. Set the dot product $\vec{v}\cdot {\vec{v}}_0=0$ to satisfy the perpendicular condition: $(v_0\cos \theta {)}^2+(v_0\sin \theta -gt)(v_0\sin \theta )=0$.
3. Expand the equation to get $v_0^2{\cos }^2\theta +v_0^2{\sin }^2\theta -v_{0}gt\sin \theta =0$.
4. Simplify using the identity ${\sin }^2\theta +{\cos }^2\theta =1$ to obtain $v_0^2-v_{0}gt\sin \theta =0$.
5. Solve for $t$ to find $t=\frac{v_0}{g\sin \theta }$.
6. Hence the answer is (A).