Question

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

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

Answer 1

The correct option is A $10\sqrt{5}m/s$

$\text{Step 1: Height and Range of Projectile}$ $\text{[Ref. Fig. 1]}$

Maximum height of the projectile: $H=\frac{u^2{\sin }^2\theta }{2g}$$=\frac{(20\sqrt{2}{)}^2{\sin }^2{45}^o}{2\times 10}=20m$

Range of the projectile: $R=\frac{u^2\sin 2\theta }{g}$$=\frac{\left(20\sqrt{2}{)}^2\sin \right({45}^o\times 2)}{10}=80m$

$\text{Step 2: Displacement at Point P}$ $\text{[Ref. Fig. 2]}$

Displacement $=OP=\sqrt{H^2+{\left(\frac{R}{2}\right)}^2}$

$\therefore$ $OP=\sqrt{{20}^2+{40}^2}=\sqrt{2000}m$

$\text{Step 3: Average Velocity Calculation}$

Average velocity of stone during its motion from starting point to its maximum height is given by:

$<V>=\frac{\text{Total Displacement(OP)}}{\text{time}\left(t_1\right)}$

Time of flight: $T=\frac{2u\sin \theta }{g}$$=\frac{2\times 20\sqrt{2}\sin {45}^o}{10}$$=4s$

Time to reach point $P{t}_1=\frac{T}{2}=2s$

$<V>=\frac{\sqrt{2000}}{2}=10\sqrt{5}m/s$

Hence average velocity from starting point to its maximum height is $10\sqrt{5}m/s$

Hence, Option $A$ is correct.