A particle is thrown with velocity u at an angle α from the horizontal. Another particle is thrown with the same velocity at an angle α from the vertical. The ratio of times of flight of two particles will be

The correct option is tan α : 1.

To find the ratio of the times of flight of the two particles, we can analyze the motion of each particle individually using the standard projectile equations.

1. Time of Flight for the First Particle:
The first particle is projected with a velocity $u$ at an angle $\alpha$ from the horizontal.
The standard formula for the time of flight ($T$) of a projectile launched from the ground at an angle $\theta$ with the horizontal is:

$T=\frac{2u\sin \theta }{g}$

For the first particle, the launch angle from the horizontal is $\mathrm{��}=\alpha$. Therefore, its time of flight ($T_1$) is:

$T_1=\frac{2u\sin \alpha }{g}$

2. Time of Flight for the Second Particle:
The second particle is projected with the same velocity $u$ at an angle $\alpha$ from the vertical.
To use the standard formula, we need the angle of projection with respect to the horizontal. Since the vertical and horizontal axes are perpendicular (at 90°), the launch angle from the horizontal is:

${\theta }_2=90°-\alpha$

Substituting this angle into the time of flight formula gives:

$T_2=\frac{2u\sin (90°-\alpha )}{g}$

Using the trigonometric identity $\sin (90°-\alpha )=\cos \alpha$, we simplify this to:

$T_2=\frac{2u\cos \alpha }{g}$

3. Calculating the Ratio of Times of Flight:
Now, we find the ratio of the time of flight of the first particle to that of the second particle:

$\frac{T_1}{T_2}=\frac{\left(\frac{2u\sin \alpha }{g}\right)}{\left(\frac{2u\cos \alpha }{g}\right)}$

Simplifying the fraction by canceling the common terms $\frac{2u}{g}$ in both the numerator and denominator:

$\frac{T_1}{T_2}=\frac{\sin \alpha }{\cos \alpha }=\tan \alpha$

Thus, the ratio of their times of flight is $\tan \alpha :1$.