Question

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

• A. $\frac{\sqrt{5}u}{2}$
• B. $\frac{\sqrt{5}u}{2\sqrt{2}}$
• C. $\frac{5u}{2\sqrt{2}}$
• D. $\frac{\sqrt{2}u}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{\sqrt{5}u}{2\sqrt{2}}$

The range is maximum when a projectile is fired with angle ${45}^0$ with the horizontal (angle of projection), hence to find the average velocity we need to things $1>$ displacement during ascent, $2>$ time of ascent

using average velocity formula

$Averagevelocity=\frac{Displacement}{Time}$

Average velocity $=\frac{D}{T/2}=\frac{2D}{T}=\frac{2\sqrt{H^2+(\frac{R}{2}{)}^2}}{T}$

Since the value of $H=\frac{u^2}{4g}$, $R=\frac{u^2}{g}$, $T=\frac{\sqrt{2}u}{g}$ (for angle ${45}^0$)

$=\frac{2\sqrt{(\frac{u^2}{4g}{)}^2+(\frac{u^2}{2g}{)}^2}}{\frac{\sqrt{2}u}{g}}=\frac{\frac{\sqrt{5}{u}^2}{2g}}{\frac{\sqrt{2}u}{g}}=\frac{\sqrt{5}u}{2\sqrt{2}}$