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 $\left(g=10\, m \,s ^{-2}\right)$

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

Answer 1

Answer: (B)

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