Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Two balls A and B are thrown with speed $u$ and $\dfrac{u}{2}$ respectively. Both the balls cover the same horizontal distance to the plane of projection. If the angle of projection of the ball B with the horizontal is $15{}^\circ$, then the angle of projection of A is:(A) ${{\sin }^{-1}}\dfrac{1}{8}$ (B) $\dfrac{1}{2}{{\sin }^{-1}}\dfrac{1}{8}$ (C) $\dfrac{1}{3}{{\sin }^{-1}}\dfrac{1}{8}$ (D) $\dfrac{1}{4}{{\sin }^{-1}}\dfrac{1}{8}$

Last updated date: 02nd Aug 2024
Total views: 67.8k
Views today: 0.67k
Verified
67.8k+ views
Hint The given question presents us with a problem of projectile motion. We have been given the initial velocities in two given cases and the angle of projection in one case. To relate the two given cases, we have been told that the range in both cases is the same. We can assume the unknown angle of projection to have a certain value and form expressions for the range in both cases and then solve the obtained expression to get the unknown value.
Formula Used: $R=\dfrac{{{u}^{2}}\sin 2\alpha }{g}$

Let’s assume that the ball A is thrown at an angle $\theta$ with the horizontal
We have been told that the initial velocity of ball A is $u$
The expression for the range in a projectile motion is given as $\left( R \right)=\dfrac{{{u}^{2}}\sin 2\alpha }{g}$ where $\alpha$ is the angle at which the object is thrown, $u$ is the velocity at which the object is thrown and $g$ is the acceleration due to gravity
Substituting the value for the given case of ball A, we get the range as $\left( {{R}_{1}} \right)=\dfrac{{{u}^{2}}\sin 2\theta }{g}$
We know that the ball B is thrown at an angle of $15{}^\circ$ with the horizontal with an initial velocity of $\dfrac{u}{2}$
The range in case of ball B can hence be given as
\begin{align} & \left( {{R}_{2}} \right)=\dfrac{{{\left( {}^{u}/{}_{2} \right)}^{2}}\sin \left( 2\times 15{}^\circ \right)}{g} \\ & \Rightarrow \left( {{R}_{2}} \right)=\dfrac{{{u}^{2}}\sin \left( 30{}^\circ \right)}{4g} \\ \end{align}
Since we have been told that the range is similar for both balls A and B, we can equate the two ranges obtained and find the unknown angle as follows
\begin{align} & {{R}_{1}}={{R}_{2}} \\ & \Rightarrow \dfrac{{{u}^{2}}\sin 2\theta }{g}=\dfrac{{{u}^{2}}\sin \left( 30{}^\circ \right)}{4g} \\ & \Rightarrow \sin 2\theta =\dfrac{\sin \left( 30{}^\circ \right)}{4} \\ & \Rightarrow \sin 2\theta =\dfrac{1}{8}\left[ \because \sin \left( 30{}^\circ \right)=\dfrac{1}{2} \right] \\ & \Rightarrow 2\theta ={{\sin }^{-1}}\dfrac{1}{8} \\ & \Rightarrow \theta =\dfrac{1}{2}{{\sin }^{-1}}\dfrac{1}{8} \\ \end{align}

From the above calculations and the obtained value of the projection angle for the ball A, we can say that the correct option for the given question is (B).

Note
To find the unknown angle, we had to substitute the value of $\sin \left( 30{}^\circ \right)$, hence the students should have a basic knowledge of the concepts of mathematics such as trigonometry. Students should also know that when sine or any other trigonometric function is transposed to the other side, it transforms into an inverse trigonometric function such as observed in the given question.