Question

A projectile can have the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

• A. $\frac{1}{R^2}$
• B. $\frac{1}{R}$
• C. $R$
• D. $R^2$

Answer 1

The correct option is C $R$

Let the two angles be ${\theta }_1$ and ${\theta }_2$. The ranges are equal when ${\theta }_1=90-{\theta }_2$
Now,
$R=\frac{u^2\sin 2{\theta }_2}{g}=\frac{u^2\sin 2{\theta }_1}{g}=\frac{2u^2\sin {\theta }_1\cos {\theta }_1}{g}$
$T_1=\frac{2u\sin {\theta }_1}{g}$
$T_2=\frac{2u\sin {\theta }_2}{g}=\frac{2u\cos {\theta }_1}{g}$
$T_1{T}_2=\frac{4u^2\sin \theta \cos \theta }{g}=2R$
Note: Arriving at ${\theta }_1=90-{\theta }_2$
Intuitively range increases from 0${}^o$ to 45${}^o$ and decreases from 45${}^o$ to 90${}^o$.
Hence ${\theta }_1$ and ${\theta }_2$ should be symmetric about 45 degrees.
Mathematically
The expression for range is $R=\frac{u^2\sin 2\theta }{g}$ where $\theta$ is between 0 and 90 degrees. Hence $2\theta$ will vary from 0 to 180 degrees.
For the range to be equal, $\sin 2{\theta }_1=\sin 2{\theta }_2$. Now, between $2\theta =0$ and $2\theta =90$, $sin2\theta$ is continuously increasing and cannot have same values for two different angles. The only possible ${\theta }_1$ and ${\theta }_2$ are so that $2{\theta }_1=180-2{\theta }_2$ (Note that $\sin \theta =\sin (180-\theta )$).