Question

A particle is projected from a horizontal plane with a velocity of $8\, \sqrt{2} \,m s ^{-1}$ at an angle $\theta$. At highest point its velocity is found to be $8 \,m \,s ^{-1}$. Its range will be $\left( g =10 m s ^{-2}\right)$

• A. 3.2 m
• B. 4.6 m
• C. 6.4 m
• D. 12.8 m

Answer 1

Answer: (D)

At highest point, velocity of projectile, $v=u \cos \theta$
$\therefore \quad 8=8 \sqrt{2} \cos \theta$
or $\cos \theta=\frac{1}{\sqrt{2}}$ or $\theta=45^{\circ}$
Range $=\frac{u^{2} \sin 2 \theta}{g}$
$=\frac{(8 \sqrt{2})^{2} \sin 2 \times 45^{\circ}}{10}$
$=\frac{64 \times 2}{10}=12.8 \,m$