Question

Find the average velocity of a projectile between the instants it first crosses half the maximum height and then again does the same. It is projected with a speed $u$ at an angle $\theta$ with the horizontal. (no need to calculate, just work it out mentally!).

• A.
• B.
• C.
• D. data insufficient

Answer 1

The correct option is A

If we focus on the part of motion after the projectile has crossed half the projectile has crossed half the maximum height and has velocity $v$ and makes angled with the x-axis.

$Averagevelocity$ = $\frac{totaldisplacement}{totaltime}$

Total displacement is happening on the x-axis only

Which if I consider this part of motion as a new projectile will be its range I know range is nothing but Horizontal component of velocity $\times$ Time period

$R=vcos\alpha \times T$

Time period in this case will be $\frac{2vsin\alpha }{g}$

So $R=vcos\alpha \times \frac{2vsin\alpha }{g}$

Now $v_{avg}$ = $\frac{R}{T}=\frac{vcos\alpha \times \frac{2vsin\alpha }{g}}{\frac{2vsin\alpha }{g}}=vcos\alpha$ = $ucos\theta$.