A projectile is fired from the origin with an initial velocity v0 at an angle theta with the horizontal. At what time will the velocity vector be perpendicular to the initial velocity vector?

A. v_0g — The velocity vector at time t is v = (v0 * cos(theta)) * i + (v0 * sin(theta) - g * t) * j. For the dot product with the initial velocity to be zero, we solve (v0 * cos(theta)) * (v0 * cos(theta)) + (v0 * sin(theta) - g * t) * (v0 * sin(theta)) = 0.

JEE Main · PhysicsHard

$A projectile is fired from the origin with an initial velocity v_{0} at an angle θ with the horizontal. At what time will the velocity vector be perpendicular to the initial velocity vector?$

A. $\frac{v _{0}}{g s i n θ}$
B. $\frac{v _{0}}{g c o s θ}$
C. $\frac{v _{0} t a n θ}{g}$
D. $\frac{v _{0}}{g t a n θ}$

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Correct answer: A — $\frac{v _{0}}{g s i n θ}$

Solution

The velocity vector at time t is v = (v_{0} cos θ) \hat{i} + (v_{0} sin θ - g t) \hat{j} . For the dot product with the initial velocity v_{0} to be zero, we solve (v_{0} cos θ) (v_{0} cos θ) + (v_{0} sin θ - g t) (v_{0} sin θ) = 0, which simplifies to t = \frac{v _{0}}{g s i n θ} .

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A projectile is fired from the origin with an initial velocity v0​ at an angle θ with the horizontal. At what time will the velocity vector be perpendicular to the initial velocity vector?

Solution

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$A projectile is fired from the origin with an initial velocity v_{0} at an angle θ with the horizontal. At what time will the velocity vector be perpendicular to the initial velocity vector?$