A car starts from rest and accelerates at $5 \mathrm{m}/{\mathrm{s}}^2$. At t = 4s, 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 $\text{'}\mathrm{\theta }\text{'}$ to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $\mathrm{\theta }$, is then given by:

1. $\mathrm{\theta }={\sin }^{-1}{\left(\frac{{\mathrm{\pi }}^2\mathrm{R}}{{\mathrm{gT}}^2}\right)}^{1/2}$

2. $\mathrm{\theta }={\sin }^{-1}{\left(\frac{2{\mathrm{gT}}^2}{{\mathrm{\pi }}^2\mathrm{R}}\right)}^{1/2}$

3. $\mathrm{\theta }={\cos }^{-1}{\left(\frac{{\mathrm{gT}}^2}{{\mathrm{\pi }}^2\mathrm{R}}\right)}^{1/2}$

4. $\mathrm{\theta }={\cos }^{-1}{\left(\frac{{\mathrm{\pi }}^2\mathrm{R}}{{\mathrm{gT}}^2}\right)}^{1/2}$