Question:
Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed $u$ at an angle $\theta$ with the horizontal
Solution:
Let particle reaches from $A$ to $B$ in time t.
By symmetry, $A B$ line is horizontal
So, displacement $A B=u_{x} t$
Velocity $=\stackrel{\text { displacement }}{\text { time }}=\frac{\text { uxt }}{t}$
$\mathrm{V}_{\mathrm{avg}}=\mathrm{u} \cos \theta$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.