Rain is falling vertically downward with a speed of 35 m/s. Wind starts blowing after some time with a speed of 12 m/s in East to West direction. The direction in which a boy standing at the place should hold his umbrella is:
1. \(tan^{-1}\Big(\frac{12}{37}\Big)\) with respect to rain
2. \(tan^{-1}\Big(\frac{12}{37}\Big)\) with respect to wind
3. \(tan^{-1}\Big(\frac{12}{35}\Big)\) with respect to rain
4. \(tan^{-1}\Big(\frac{12}{35}\Big)\) with respect to wind

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 \(\mathrm{R}\) with a uniform speed takes a time \(\mathrm{T}\) 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 \(\mathrm{4R}.\) The angle of projection, \(\theta\) 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. $\mathrm{\theta }={\cos }^{-1}{\left(\frac{{\mathrm{\pi }}^2\mathrm{R}}{{\mathrm{gT}}^2}\right)}^{1/2}$