# Average torque on a projectile

1 vote
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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 ?

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

1 vote

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 $\frac12mu^2\sin2\theta$.

answered Feb 1, 2017 by (28,806 points)
edited Feb 2, 2017
ohh! I didn't solve that ahead