Question

From a point on the ground particle is projected with initial velocity u, such that its horizontal range is maximum. the magnitude of average velocity during its ascent

• A. $\frac{\sqrt{5}u}{\sqrt{2}}$
• B. $\frac{5u}{4}$
• C. $\frac{\sqrt{3}u}{2\sqrt{2}}$
• D. none

Answer 1

The correct option is D none

Let's consider, $u=$ initial velocity of the projectile

$\theta ={45}^0$ for maximum horizontal range

Horizontal range,

$R=\frac{u^{2}sin2\theta }{g}=\frac{u^{2}sin{90}^0}{g}$

$R=\frac{u^2}{g}$

Maximum height,

$H=\frac{u^{2}si{n}^2\theta }{2g}=\frac{u^{2}si{n}^2{45}^0}{2g}$

$H=\frac{u^2}{2g}(\frac{1}{\sqrt{2}}{)}^2$

$H=\frac{u^2}{4g}$

Time of flight, $T=\frac{2usin\theta }{g}$

$T=\frac{2usin{45}^0}{g}=\frac{2u}{g}\frac{1}{\sqrt{2}}$

$T=\frac{\sqrt{2}u}{g}$

Applying Pythagoras theorem In the above figure,

$D=\sqrt{H^2+(R/2{)}^2}$

$D=\sqrt{(\frac{u^2}{4g}{)}^2+(\frac{u^2}{2g}{)}^2}$

$D=\frac{\sqrt{5}{u}^2}{4g}$

Average speed during ascent,

$v_{avg}=\frac{D}{T/2}=\frac{2D}{T}$

$v_{avg}=2\frac{\frac{\sqrt{5}{u}^2}{4g}}{\frac{\sqrt{2}{u}^2}{g}}$

$v_{avg}=\frac{\sqrt{5}u}{2\sqrt{2}}$

The correct option is D.