Question

When a particle is projected at some angle to the horizontal, it has a range?

Answer 1

let the angle is @.and initial velocity=u
so range = u^2sin2@/g

Answer 2

The kinematic equations of the motion of a particle  are as follows assuming the center of the coordinate system is the initial position of the particle

                      $x=v_o cos\alpha \times t\\y=v_o sin\alpha \times t-(gt^2)/2$

Here we can see that x-axis is in the direction towards the motion of the particle. Y-axis is directed upwards.

So the condition for calculating the time of fall is y = 0

                     $v_o sin\alpha \times -(gt^2)/2=0\\t_o=(2v_o sin\alpha)/g$

So the range is R=x(t_o )=v_o cos\alpha \times t_o

                    $=v_o \times \sqrt(1-sin^2\alpha ) \times t_o\\=v_o \sqrt{(1-(\frac{(gt_o)}{(2v_o )})^2 )} \times t_o\\=t_o/2 \sqrt{(4v_o^2-g^2 t_o^2 )}$

Squaring we get

                    $R^2=\frac{(t_o^2)}{4} (4v_o^2-g^2 t_o^2 )\\v_o^2= \frac{R^2}{(t_1^2 )} + \frac {(g^2 t_1^2)}{4}\\=\frac{R^2}{(t_2^2 )} + \frac {(g^2 t_2^2)}{4}\\\frac{R^2}{(t_1^2 )} - \frac {R^2}{(t_2^2 )} = \frac {(g^2 t_2^2)}{4} - \frac{(g^2 t_1^2)}{4}$

                     $t_1 t_2=2R/g$

Hence Answer is option 3