Question

The point from where a ball is projected is taken as the origin of the coordinate axes. The x and y components of its displacement are given by $x = 6t$ and $y = 7t - 5t^{2}$ .what is the angle of projection?

• A. $tan^{-1} \left(\frac{1}{4}\right)$
• B. $tan^{-1} \left(\frac{4}{3}\right)$
• C. $tan^{-1} \left(\frac{1}{8}\right)$
• D. $tan^{-1} \left(\frac{1}{6}\right)$

Answer 1

Answer: (B)

tan $\theta=\frac{v_{y}}{v_{x}}=\frac{8}{6} =\frac{4}{3} \,$ or $\,\theta=tan^{-1} \frac{4}{3}$