Question

A projectile is projected with initial velocity

$(6\widehat{i} + 8\widehat{j})m/sec.$ If g = 10 ms^–2, then horizontal range is

• A. 4.8 metre
• B. 9.6 metre
• C. 19.2 metre
• D. 14.0 metre

Answer 1

Answer: (B)

9.6 metre

Answer 2

Initial velocity $= \left( 6\widehat{i} + 8\widehat{J} \right)m/s$ (given)

Magnitude of velocity of projection $u = \sqrt{u_{x}^{2} + u_{y}^{2}}$

$= \sqrt{6^{2} + 8^{2}}$= 10 m/s

Angle of projection $\tan\theta = \frac{u_{y}}{u_{x}} = \frac{8}{6} = \frac{4}{3}$ ∴ $\sin\theta = \frac{4}{5}$and $\cos\theta = \frac{3}{5}$

Now horizontal range $R = \frac{u^{2}\sin 2\theta}{g}$ $\sqrt { \frac { r F } { m } }$

$= \frac{(10)^{2} \times 2 \times \frac{4}{5} \times \frac{3}{5}}{10} = 9.6meter$