Question

A projectile is projected with initial velocity $(6\hat{i}+8\hat{j})m/sec.Ifg=10m{s}^{-2}$, then horizontal range is

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

Answer 1

The correct option is B

9.6 metre

Initial velocity $(6\hat{i}+8\hat{J})m/s$ given

Magnitude of velocity of projection $u=\sqrt{u_x^2+u_y^2}=\sqrt{6^2+8^2}=10m/s$

Angle of projection $tan\theta =\frac{u_y}{u_x}=\frac{8}{6}=\frac{4}{3}\therefore sin\theta =\frac{4}{5}andcos\theta =\frac{3}{5}$

Now horizontal range $R=\frac{u^{2}sin2\theta }{g}=\frac{u^{2}2sin\theta cos\theta }{g}=\frac{(10{)}^2\times 2\times \frac{4}{5}\times \frac{3}{5}}{10}=9.6meter$