Question

Three particles $\text{A},\text{B}$ and $\text{C}$ are projected from the same point with the same initial speeds making angles ${30}^{\circ },{45}^{\circ }$ and ${60}^{\circ }$ respectively with the horizontal. Which of the following statements is correct ?

• A. $\text{A},\text{B}and\text{C}$ have unequal ranges.
• B. Ranges of $\text{A}$ and $\text{C}$ are equal and less than that of $\text{B}$.
• C. Ranges of $\text{A}$ and $\text{C}$ are equal and greater than that of $\text{B}$.
• D. $\text{A},\text{B}$ and $\text{C}$ have equal ranges.

Answer 1

The correct option is B Ranges of $\text{A}$ and $\text{C}$ are equal and less than that of $\text{B}$.

Here $\text{A}$, $\text{B}$ and $\text{C}$ have same initial speed $u$ but different projection angle from horizontal.

For a projectile with initial speed $u$ and projection angle $\theta$ from the horizontal range is given by

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

In this case, $u$ is same

If we take $u$ equal then $R$ depends on $\sin 2\theta$.

For maximizing $R$,

$\sin 2\theta =1\Rightarrow \theta ={45}^{\circ }$

So, Out of $\text{A}(\theta ={30}^{\circ })$, $\text{B}(\theta ={45}^{\circ })$ and $\text{C}(\theta ={60}^{\circ })$

$\text{B}$ will have maximum Range.

Now, if range is same for two different angles,

$\Rightarrow \sin 2{\theta }_1=\sin 2{\theta }_2$

$\because {\theta }_1\ne {\theta }_2$

so we take,

$2{\theta }_1=\pi -2{\theta }_2\Rightarrow {\theta }_1+{\theta }_2=\pi /2$

For $\text{A}$ & $\text{C}$,

${\theta }_1+{\theta }_2={90}^0$, So $R_A=R_C$

Thus, $R_A=R_C<R_B$

Hence, option $\left(b\right)$ is correct.