Question

A projectile is fired from point P at an angle of 45${}^o$ horizontally as shown below. If g is the acceleration due to gravity, then the speed required to reach a point Q lying on the horizontal surface at a distance of R from point P is

• A. $\sqrt{Rg/2}$
• B. $\sqrt{Rg}$
• C. $\sqrt{2Rg}$
• D. $\sqrt{3Rg}$

Answer 1

The correct option is B $\sqrt{Rg}$

$R=\frac{V_2^{2}Sin2\alpha }{g}$

$\text{If}\alpha ={45}^o$

$R=\frac{V_2^2\times 1}{g}$

$\therefore V_2=\sqrt{Rg}$