A car starts from rest and accelerates at $5$ $\mathrm{m}/{\mathrm{s}}^2$. At \(t=4\) s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at \(t=6\) s? (Take \(g=10\) m/s²)
1. $20\sqrt{2}$ $\mathrm{m}/\mathrm{s}, \mathrm{0 }\mathrm{m}/{\mathrm{s}}^2$
2. $20\sqrt{2}$ $\mathrm{m}/\mathrm{s},$ $10$ $\mathrm{m}/{\mathrm{s}}^2$
3. $20$ $\mathrm{m}/\mathrm{s},$ $5$ $\mathrm{m}/{\mathrm{s}}^2$
4. $20$ $\mathrm{m}/\mathrm{s},$ $0$

A particle moving in a circle of radius \(R\) with a uniform speed takes a time \(T\) to complete one revolution. If this particle were projected with the same speed at an angle \(\theta\) to the horizontal, the maximum height attained by it equals \(4R.\) The angle of projection, \(\theta\) is then given by:
1. \( \theta=\sin ^{-1}\left(\frac{\pi^2 {R}}{{gT}^2}\right)^{1/2}\)
2. \(\theta=\sin ^{-1}\left(\frac{2 {gT}^2}{\pi^2 {R}}\right)^{1 / 2}\)
3. \(\theta=\cos ^{-1}\left(\frac{{gT}^2}{\pi^2 {R}}\right)^{1 / 2}\)
4. \(\theta=\cos ^{-1}\left(\frac{\pi^2 {R}}{{gT}^2}\right)^{1 / 2}\)