Average torque on a projectile

Question

Average stock on a projectile of mass M initial speed u and angle of projection $\theta $ between initial and final position about the point of projection.

what I did:
$r×F=r×mg
=ucos\theta .t .mg $(because we take perpendicular component)

then what's wrong as answer is in terms of m,u, sin or cos only ?

Comments

What is $t$? What is the "correct" answer?

Answer 1

You are correct so far. You have to substitute a value for $t$, the average time during flight, ie the time taken to reach maximum height, which is $\frac{u\sin\theta}{g}$. Then average torque is $\frac12mu^2\sin2\theta$.

Comments

ohh! I didn't solve that ahead