Question

In what time the direction of final velocity of a projectile projected with velocity u at an angle \(\alpha\) be perpendicular to the direction of initial velocity?

Ans:\[\frac{u}{g \sin \alpha}\]

Answer 1

form the equation of projectile's motion.
find X(t) & Y(t) and then find Y(x).
it will be an equation of parabola.
differentiate the equation wrt X to find dy/dx.
if a is angle of projection then put
dy/dx = -1/ tan(a).
solve for t thereafter.

Answer 2

Thanks @quarkine

Answer 3

The initial velocity vector is u = ucos(a) i + usin(a) j
Let the velocity at which it becomes perpendicular to u be v.
Now, v = ucos(a) i + x j (x-component of velocity remains constant in projectile motion)
Two vectors a and b are perpendicular if a.b = 0 (dot product)
Take dot product of u and v and equate with 0
\[u^{2}\cos^{2}\alpha + u.\sin\alpha.x = 0\]
\[x = -ucos^{2}\alpha/\sin\alpha\]

Now, x = usin(a) - gt
=> t = (usin(a) - x)/g
\[t = \frac{usin\alpha + \frac{ucos^{2}\alpha}{\sin\alpha}}{g}\]
\[t = \frac{u}{gsin\alpha}\]

Answer 4

Thanks @FoolAroundMath ... nicely done buddy!