Question

A particle is projected from the ground with an initial speed u at an angle $\theta$ with the horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

• A. $ucos\theta$
• B. $\frac{u}{2}\sqrt{1+co{s}^2\theta }$
• C. $\frac{u}{2}\sqrt{1+2co{s}^2\theta }$
• D. $\frac{u}{2}\sqrt{1+3co{s}^2\theta }$

Answer 1

The correct option is D $\frac{u}{2}\sqrt{1+3co{s}^2\theta }$

The maximum height of a projectile,

$H=\frac{u^{2}si{n}^2\theta }{2g}$. . . . . . . . .(1)

Range of a projectile,

$R=\frac{2u^{2}sin\theta .cos\theta }{2g}$. . . . . . . . . .(2)

By Pythagoras theorem in $\Delta OAB$

$d=\sqrt{H^2+\frac{R^2}{4}}$. . . . . . .(3)

Time to react the maximum height $A$

$t=\frac{R}{2vcos\theta }$. . . . . . . . .(4)

Average velocity of the particle between its point of projection and highest point of trajectory is

$v_{avg}=\frac{d}{t}=2vcos\theta \sqrt{\frac{1}{4}+(\frac{H}{R}{)}^2}$.. . . . . . .(5)

Divide equation (1) by (2)

$\frac{H}{R}=\frac{tan\theta }{4}$. . . . . . . . .(6)

Substitute equation (6) in equation (5), we get

$v_{avg}=\frac{u}{2}\sqrt{1+3co{s}^2\theta }$

The correct option is D.