Question

For a given velocity, a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the time of flight in the two cases, then:

• A. $t_1{t}_2\propto R$
• B. $t_1{t}_2\propto R^2$
• C. $t_1{t}_2\propto \frac{1}{R^2}$
• D. $t_1{t}_2\propto \frac{1}{R}$

Answer 1

Answer: (A)

$t_1{t}_2\propto R$

Let the projectiles be projected at a velocity $u$.

Let the angles of projection be ${\theta }_1$ and ${\theta }_2$.

So, from the condition provided:

$R=\frac{u^2}{g}\sin 2{\theta }_1=\frac{u^2}{g}\sin 2{\theta }_2$

Since, for both the projected angles, range is same, the angles follow the relation: ${\theta }_1+{\theta }_2={90}^0$, or, ${\theta }_2={90}^0-{\theta }_1$

Now, the times of flight, $t_1$ and $t_2$ may be expressed as:

$t_1=\frac{2u\sin {\theta }_1}{g}$ and $t_2=\frac{2u\sin {\theta }_2}{g}=\frac{2u\cos {\theta }_1}{g}$

So, $t_1{t}_2=\frac{2u\sin {\theta }_1}{g}\times \frac{2u\cos {\theta }_1}{g}=\frac{2u^2}{g^2}\times \left(2\sin {\theta }_1\cos {\theta }_1\right)$$=\frac{2u^2}{g^2}\sin 2{\theta }_1=\frac{2R}{g}$

$\Rightarrow t_1{t}_2\propto R$