Question

A stone is projected with a velocity $20\sqrt{2}m{s}^{-1}$ at an angle of $45^{\circ }$ to the horizontal. The average velocity of stone during its motion from starting point to its maximum height is $\left(g=10m{s}^{-2}\right)$

• A. $5\sqrt{5}m{s}^{-1}$
• B. $10\sqrt{5}m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $20\sqrt{5}m{s}^{-1}$

Answer 1

Answer: (B)

When projectile is at $A$,
then
$OC=\frac{R}{2}=\frac{1}{2}\frac{u^2}{g}\sin 2\theta$
$=\frac{1}{2}\times \frac{(20\sqrt{2}{)}^2}{10}\sin 2\times 45^{\circ }=40m$
$AC=H=\frac{u^2{\sin }^2\theta }{2g}=\frac{(20\sqrt{2}{)}^2}{2\times 10}{\sin }^{2}45^{\circ }=20m$
$\therefore$ Displacement, $OA=\sqrt{O{C}^2+C{A}^2}=\sqrt{40^2+20^2}$
Time of projectile from $O$ to $A$
$=\frac{1}{2}\left(\frac{2u\sin \theta }{g}\right)=\frac{u\sin \theta }{g}=\frac{(20\sqrt{2})\sin 45^{\circ }}{10}=2s$
Average velocity $=\frac{\text{ displacement }}{\text{ time }}$
$=\frac{\sqrt{40^2+20^2}}{2}=10\sqrt{5}m{s}^{-1}$